Prof.Anarse.D A
Yashasvi Science Academy
Physics-II
10. Wave Theory of Light
Q.1 Write a short note on Huygens Wave Theory of light.
Ans:i.
ii.
iii.
iv.
v.
vi.
vii.
viii.
According to this theory light is propagated in the form of waves.
These waves are emitted by source of light and travel with uniform velocity in homogeneous
medium.
Different colours of light are due to the difference wavelength of light.
Huygens considers a luminiferous ether layer through which light waves can pass.
He assumes that velocity of light in rarer medium is greater than velocity of light denser medium
which is experimentally proved.
When light waves enter in our eyes, then we get sensation of light.
Reflection, Refraction, diffraction and interference can be satisfactory explain on the basis of wave
theory of light.
According to this theory, waves are longitudinal in nature hence; polarisation and photoelectric
effect cannot be explained with the help of this theory.
Q.2 Explain the concept of wave front & wave normal.
Ans:-
Consider a point source S of light situated in air, the waves emitted by the source travel in all possible
directions.Let C be the velocity of light. Therefore, each wave will cover a distance C in time t.
Draw a circle with center S & radius c such spherical surface is called as spherical.
Wave front If at t =0 source is emitted a crust, and then crust will be produce at every point on
sphere.Therefore, every point of sphericalwavefront is in the same phase.
o Wave front :The locus of all the points of the medium to which wave reaches simultaneously, so that all the
points are in same phase is called as Wave front.
o Wave normal :A Perpendicular drawn to surface of wave front at any point in the direction of propagation of light is
called as Wave normal.
Plane wave front & wave normal
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Spherical wave front & wave normal
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Q.3 State Huygens constructions & Huygens principle.
Ans:-
A. Huygens’s Construction :1) If the nature & position of the wave front at some instant is known, it is possible to determine
the nature and position of wave front at later instant by means of geometrical construction
called as Huygens constructions.
B. Huygen’s Principle:1) Every point on a wave front acts as a secondary source of light sending out secondary waves.
2) The envelope of all these secondary waves at any later instant gives the new wave front at
that instant.
Q.4 Explain Huygens construction of a spherical wave front.
Ans:
Consider a point source‘s’ situated in air medium.
Let PQRT be the points at the same distance from SThe waves emitted by the source with reach
thesePoints simultaneously.So, that points will be on theSpherical wave front.As soon as spherical
wave front is formed every point on it
Becomes secondary source & emits secondary waves.Draw spheres with the centre P,Q,R,T & radius
equal to ct where’C’ is velocity of light .Each sphere will represent a secondary wavefront.A surface
which is tangential to all spheres will represent a new wave front after time’t’.Dotted lines shows the
secondary waves in backward direction but since light is propagated in forward direction such waves
do not exist.
Q.5 Explain Huygens’s construction of plane wave front.
Ans:Consider section AB of plane wave front, suppose it is moving
From Left to right. Draw spheres with every point on AB as
centre & radius equal to ct,Where C is velocity of light.
The envelope A’B’ of these spheres will represent new Wave front
after time t. Dotted lines shows the secondary waves in
backward direction, but since light is propagated in forward direction,
such waves do not exist.
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Physics-II
Q.6 With the help of diagram explains refraction of light from a plane surface on the
basis of wave theory of light.
Ans:-
Let X.Y be the plane surface which separated air and denser medium. Suppose a plane wave front is bounded
by two rays PA and QB incident in air medium on surface XY.Initially wave front reaches to the point A
becomes secondary source of light an emits secondary waves. Let C1 be the velocity of light in rarer (air)
medium and C2 be the velocity of light in denser medium.
If incident wave front moves from point B to pt C in time t,then d(Bc) =C1t.
In the same time t, point A covers a distance ‘C2t ‘ in denser medium. Draw a tangent CD, draw AD and
produce it up to S.Draw CR II AS. Draw normal MN to surface XY.
Therefore, PAM =i =angle of incident
NAD =r =Angel of refraction
From the geometry of figure,
PAM = BAC =i
NAD = ACD =r
PAM + MAB =90
MAB + BAC=90
PAM + MAB =
MAB +
BAC
PAM = BAC
And ,also
NAD +
DAC =90
DAC +
ACD =90
NAD +
DAC = DAC + ACD
NAD = ACD
In ∆ ABC,
Sin i =
…………………….
1
In ∆ ADC ,
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Sin i =
……………………
Physics-II
2
Divide equation 1 by 2.
=
×
=
=
=
…………………
3
We know,
=n
………………...
4
…………….. refractive index
From equation 3 & 4
n=
This is called as Snell’s law
n> 1
>1
C1 > C2
Thus velocity of light in air medium is greater than velocity of light in denser medium which
supports Huygens wave theory of light.
Q.7 Explain the following term.
1. Unpolarised light(ordinary light)
Light waves are transverse electro-Magnetic wave &
In which electric & magnetic Field vectors mutually
Perpendicular to each other & them are perpendicular to
direction of Propagation.
A light in which the vibration of Electric field vectors are
in all possible direction which is perpendicular to direction
of propagation is called as unpolarised or ordinary light.
2. Plane Polarised light:-
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direction of
prorogation of light
Vibration in parallel plane
Prof.Anarse.D A
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Physics-II
3.Plane Polarized light
Phenomenon of light in which vibration of light are confined or restricted in a single plane is called
as plane polarized light.
4.Plane of Polarization:
- The plane in which there is no vibration of polarised light is called as plane
of Polarisation.
5.Plane of vibration:
- The plane in which vibrations of polarised light are present is called as plane of
vibration.
Q.8. State and explain Brewster’s law of polarization of light.
Ans: - Statement:When light is incident on refracting surface at an angel of incidence equal to angel of polarizer then the
refractive index of that medium is numerically
Equal to tangent of angel of polarizer.
Mathematically,
n = tan P.
Consider ray AB of ordinary light is incident on plane surface which separatesTwo media. The ray of light is
refracted along BD and it is reflected along BC.It is found that reflected light partially polarized of certain
angle of incidence, the reflected light is completely plane polarise, that certain angle is called as Polarizing
angle and angle of polarization, reflected ray and refracted ray are ⊥ to each other.
From the fig,
r = 90 - P
We know,
n=
n=
(
)
n=
n = tan P
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Physics-II
Q.9. what are Polaroid? How they can be constructed?
i.
ii.
Polaroid is large size polarising film which is used to obtain plane polarised light of large cross
section. It is based on principle of selective absorption.
Construction :pola
roid
Unpolarised light
plane polarised light
Polaroid
1
Polaroid 2
Unpolarised light
plane polarised light
Pola
roid
2
Polaroid
1
Plane polarized light
Unpolarised light
No Light
It is an extremely thin layer of crystalline substance called Herapothite(Iodosulphate of quinine) to prepare
a thin layer of sheet .Ultra microscopic crystal of Herapathite is embedded on thin sheet of nitrocellulose
such that the optic axis of all the crystal are parallel .
When light is incident on Polaroid. Then the vibrations parallel to the transmission plane can pass through
it & all other vibration is eliminated, thus we get plane polarized light. Two Polaroid’s P1 and P2 are
kept parallel to each other then the light transmitted by first Polaroid is transmitted by second Polaroid
.When two Polaroid’s are mutually perpendicular to each other, then the light transmitted by 1st Polaroid
is completely block by 2nd Polaroid.
Q.10 State uses of Polaroid’s.
i.
ii.
iii.
iv.
v.
vi.
It is used to obtain plane polarised light and to detect plane polarised light.
It is used in sunglass to avoid glare or sharing.
It is used as filter in photographic cameras to eliminate glare of reflected light.
It is used in spectacles to see 3-D effect in 3-D motion picture.
It is used in headlight of vehicle to avoid intense light of approaching vehicle.
They are used in calculators, watches, monitors of laptops which have LCD screen.
Q.11.Explain the Doppler Effect in light and hence explains the red and blue shift.
Ans:- The Doppler effects is a symmetric i.e it depends only on the relative velocity of the source
and
observer ,irrespective of which of the two is moving. The difference occurs because sound require a material
medium and the speed of the source and observer are measured relative to the medium .light does not
require a medium for propagation and the speed of light is same for any observer and /or the source is
moving, the formula for frequency of light obtained by using the theory of relativity is
± /
V’=v [
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(
/ )
]
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When vr<<c
V’=v{1 ± }
Red and blue shift
Due to Doppler Effect a wavelength in the middle of the visible spectrum will be shifted towards
red when source and the observer move away from each other and toward blue, if they approach
each other. The term red and blue shift are used for frequency increases (wavelength decreases) and
frequency decreases (wavelength increases) respectively. The measurement of Doppler shift helps
to study the motion of the star and galaxies.
Important Formulae
1)
n=
ng = = = 2)
a
3)
wave no =
4)
n = tan P
5)
r = 90-P
Q.12. Numericals :
1) Light is incident on glass slab making an angle of 600 with its surface .Calculate angle of
refraction at velocity of light in glass if refraction index of glass is 1.5 & velocity of light in air is 3
×108 m/s
Ans- i =900-600 =300
ang =1.5
Ca =3 ×108 m/s
To find - i) r ii)Cg
i)
n = Sin r =
=
=
.
× . sin r =
sin r =0.3333
r = sin-1 (0.3333)
r= 190 28’
ii)
a
ng =
1.5 =
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×
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×
Cg = . Cg =2× 108 m/s
2) Ray of light is incident on glass slab of refractive index 1.5 making an angle of 400 with surface.
find angle of refraction in glass & speed of light in glass. Speed of light in air is 3×108 m/s
Ans:- Given data ang =1.5 Cg =3× 108 m/s
i= 900-400 =500
to find:-i) r=?
ii) Cg=?
We know ,n = i)
sin r =
=
.
.
=
.
r = sin-1 (0.5107)
=300,41’
We know,
ang =
ii)
Cg =
=
×
. Cg =2× 108 m/s
3) Wavelength of green light in air is 5300A0 .Determine velocity and wavelength of this light in
glass of refractive index 1.5. Velocity of light in air 3×108 m/s
Ans:Λa = 5300A0.
ang = 1.5
Ca = 3× 108 m/s
To find i) Cg ii ) λg
i)
ang =
×
1.5 = Cg =2 ×108 m/s
i)
ang =
. 1.5 = λg =
.
=
53000
15
λg =3533.33 A0
4) If refractive indices of diamond & glass are 2.4 & 1.5 respectively .Compare the velocities of light
in diamond & glass.
a nd = 2.4 and
a ng =1.5
to find a
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nd = = --------1
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Prof.Anarse.D A
a
ng = Yashasvi Science Academy
Physics-II
----------2
divide equation 2/1 we get
= × 1.5
C
=
2.4
C
C
5
=
C
8
5) The refractive index of water is 1.33 .what will be the polarising angle for water surface.
Given,
N=1.33
P=?
According to Brewster’s law,
n = tan P
1.33 =tan P
p =tan-1 (1.33)
p=5304’
6) What should be the angle of incidence for a glass plate of refractive index 1.5.so that the
reflected ray is completely polarised.
Ans:- If polarized ray is completely polarized i=p.
By according to Brewster’s law,
n =tan P
1.5 =tan P
P= tan-1 (1.5)
P =560 19’
i=P
i = 560 19’
7) If a glass plate of refractive index 1.7321 is to be used as a polariser. What should be the
polarising angle & angle of refraction?
Ans :- n =1.7321
We know p=i
i=600
To find r =? ,p =?
n = = We know n=tan p
1.7321=tan p
P=tan-1(1.7321)
P=600
I=60
sin r =
√
× .
=
.
.
r= sin-1(0.5000)
r= 300
8) The refractive indices of water for red & violet colours are 1.325 and 1.334 resp. Find the
difference between the velocities of rays for these two colours in water.
Ans:- nr =1.325
nv =1.334
nr < nv
Cr > Cv
To find Cr - Cv = ?
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nr =
Yashasvi Science Academy
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i.e.
Cr = and nv = Cv =
Consider, Cr – Cv = − −
=Ca
= Ca
= Ca
= 3×10
=
.
.
−
.
.
.
× .
.
.
.
8
.
× .
×
× .
. ×
= . × .
Log (Cr – Cv) =log 2.7 –log 1.325 + log 1.334
= 0.4314 -0.1222 +0.1242
= 0.4314 -0.2464
= 0.1840
(Cr – Cv) = 1.528×106
9) Red light of wavelength 6400 A0 in air has wavelength 4000 A0 in glass. If the wavelength of violet
light in air is 4400 A0.What is its wavelength in glass.
Ans - λr(air) =6400 A0
λr(glass) = 4000 A0
λv(air) = 4400A0
λv(glass) = ?
Refractive index of red & violet color is approximately same.
i.e. n =
(
)
(
=
(
=
)
(
(
)
)
×
λv(glass) =
=
)
λv(glass) =2750 A0
10) If the difference in velocities of light in glass & water is 0.25×108m/s . Find the velocity of
light in air refractive index of glass is 1.5 & refractive index of water is
Ans :- ng =1.5 =
=
nw =
Ca =?
ng >nw
Cg < Cw
ng = Page | 10
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Cg = And nw =
Cw = Cw –Cg = −
= Ca
= Ca
− /
= Ca
−
/
−
= Ca 0.25 ×108 = Ca ×
Ca = 0.25×12 ×108
Ca = 3×108m/s
10) If the critical angle of medium is sin-1( ).Find the Polarising angle.
Ans :- C = sin-1( )
P= ?
Sin C =
We know, n =
=
/
=
n= 1.6667
we know , n = tan P
1.6667 = tan P
P =tan-1 (1.6667)
P = 5902’
11. Interference & Diffraction
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Q.1 State the principle of superposition of waves.
Ans :- If two waves, travelling through a medium, arrived at a point simultaneously, each wave produces
its own displacement at that point independent of other. The resultant displacement at that point is equal
to vendor sum of displacement due to two waves.
Q.2 Define the term Interference.
Ans :- The phenomenon of enhancement or cancellation of displacement produce due to superposition of
waves is called as interference.
Q.3 State the condition for steady interference pattern.
i.
The two sources of light must be coherent.
ii.
The two sources of light must be monochromatic.
iii.
The two sources of light must be equally bright.
iv.
The two sources of light must be narrow.
v.
The two sources of light must be close to each other.
Q.4 .why two sources of light must be coherent to obtain steady interference pattern?
Ans:- Two sources are coherent sources if phase difference between them is either zero or constant.
If two different sources of light having equal wavelength & equal amplitude are used ,then
also steady interference pattern cannot be obtained because waves form by these sources undergo rapid &
irregular changes of phase ,so that intensity at any point is never constant. Hence....
 Production - sources which are derive from one & same original source is called as coherent source.
A source & its image can also serve as coherent source.
Q.5 why two sources of light should be monochromatic to obtain steady inherence pattern?
Ans:- A monochromatic source of light is a source which emits light waves of only one wavelength .If
monochromatic light is used sharp& well defined interference pattern is obtained .Hence...
Q.6 why two sources of light must be equally bright to obtain steady interference pattern?
Ans :- Sources are equally bright if they emit waves of equal amplitude.
Intensity of light is directly proportional to square of amplitude .If amplitude are not equal the point
at which waves arrive out of phase will not be completely dark therefore interference pattern consist of
bright & less bright points. Hence....
Q.7 why two sources should be narrow to obtain steady interference pattern?
Ans: - wide source is equivalent to no of narrow sources Each source emit wave which interfere with each
other .So that number of wave trains are obtained instead of two .Therefore resulting interference pattern
will not be sharp & clear. Hence....
Q.8 why two sources should be close to each other to obtain steady interference pattern?
Ans:- Two sources should be closed to each other to get sufficient separation between regions of brightness
& darkness .when two sources are far away, Bright & Dark regions are very close to each other & they
cannot be clearly observed. Hence....
Q.9 Obtain a condition to get constructive & destructive interference.
Ans: - 1.When two waves are in phase, then we get constructive interference (bright point) .If waves are in
phase difference between them =0, 2Π, 4Π, 6Π, But phase difference 2Π corresponds to path difference λ
path diff = 0,λ ,2λ , 3λ... nλ ,where n=0,1,2,3...
Thus, there is a constructive interference if path difference between two waves arriving at the pt is an
integral multiple of wavelength λ.
2. When two waves are out of phase then there is destructive interference (dark pointt) if waves are out of
phase, phase difference between them =π, 3π, 5π, 7π...
But phase difference 2π corresponds to path diff λ
Path diff = , , , ,....
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=
−
Yashasvi Science Academy
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λ, where n=1, 2, 3...
=(2 − 1) , where n=1, 2, 3....
Thus, there is a destructive interference if the path difference between
Two waves arriving at a point is an odd multiple of half the wavelength.
.10.Give the theory of interference pattern & hence obtain an expression for band width (fringe
width).
Ans: - Consider, Two coherent monochromatic Sources A & B Separated by a distance‘d’.It produces
Interference pattern on screen place at distance D from The sources. Let, O be the mid-point of AB draw
AM,DP, BN Perpendicular to screen. Let λ be the wavelength of light emitted by each Sources. Let Q be a
point on the screen at a distance x
From point P Joint BQ and AQ.
∴ Path difference between the waves of light reaching the
Pt. Q is BQ-AQ
In ∆ AMQ,
∴ AQ2 = AM2 +MQ2
AQ2 =D2+ −
In ∆BNQ,
BQ2 = BN2 +NQ2
+
BQ2 =D2+
Consider, BQ2 – AQ2=
=
+
+
+
+
+
+ +
+
= 2xd
(BQ-AQ) (BQ+AQ) =2xd
BQ-AQ =(
)
=
+
−
-
+
+
-
−
+
+
−
BQ2-AQ2
∴X& d are very small compare to D
Let BQ-AQ =D
∴ BQ-AQ =
=
∴ BQ-AQ =
Point Q will be bright p if
Point Q will be dark p if
 Band width (Fringwidth) :Page | 13
= nλ where n =0, 1, 2,...
=
−
λ where n=1, 2, 3,
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Distance between two consecutive bright bands or dark bands is called as band width or fringwidth
.
I.
To find the distance between two consecutive bright bands
th
Let n bright band is at a distance xn from central band &
(n+1)th bright band is at a distance
Xn+1 from central band.
Let x= xn+1 -xn
For nth bright band
= nλ -----------1
For (n+1)th bright band
= (n+1) λ ---------2
∴
(
∴
-
−
=(n+1)λ –nλ
…………… (2-1)
) = nλ+λ-nλ
( )=λ
∴ x=
. To find the distance between two consecutive dark bands:
Let mth dark band is at a distance xm from central band & m+1th dark band is at distance xm+1
From central band,
Let x =
−
For mth dark band,
=
−
---------3
th
For (m+1) dark band,
.
=
+1−
.
∴
.
∴
(
=
−
−
+
.
-------4
=
+
−
−
) = mλ + λ - mλ + λ
( x ) =λ
X=
Thus distance between two consecutive bright band=distance between two consecutive dark band.
Hence, dark band & bright bands are equally spaced
.
Q.11. Describe Young’s double slit Experiment.
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Young perform the experiment to demonstrate interference of light .A pin hole S is illuminated by sunlight.
This light is along to form on two more pin hole S1 & S2 kept closed to each other .These sources acts as a
coherent source of light & produces steady interference pattern on screen which consist of alternate dark &
bright bands.
Sunlight is replaced b monochromatic source of light & pinholes are replacing by narrow
slits then also interference pattern consist of alternate dark & bright bands. Sharp & clear interference
pattern is obtained .These bands are called as interference bands or fringes & this experiment is called as
Young’s double slit experiment.
To calculate wavelength λ :Wavelength of monochromatic source of light can be calculated by using the formula λ = .
Where D=distance between slit &screen
d= distance between two slits
x= distance between two consecutive bright bands or dark bands.
Important of Young’s experiment:i.
It was the 1st experiment to demonstrate interference of light.
ii.
It concludes that light is propagated in the form of waves which supports Huygens’s wave theory of
light.
iii.
Wavelength of monochromatic source of light can be calculated by using the formula λ =
Q.12 Describe Biprism Experiment for determining wavelength of monochromatic light.
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Ans: - Diagram
Biprism is a prism having very small refracting angle narrow slit ‘S’ is eliminated by monochromatic light
which is placed parallel to refracting edge of biprism B.The rays are refracted from prism which further
passes through 2 vertical images S1& S2 of slits Interference pattern occurs in shaded region where light
waves from S1& S2 overlap to each other. The interference fringes can be seen through micrometre
eyepiece E.
In the biprism experiment slit, biprism & eyepiece are mounted on optical bench by
adjustable stands. Optical bench is very heavy and about 1.5 m length. Scale is marked on optical bench.
Initially, slit biprism & eyepiece are kept at same height. The slit is eliminated by
monochromatic source of light. The slit is very narrow & vertical. Biprism is move along horizontal axis
sharp & clear interference pattern is observed through eyepiece.
 Determine of Wavelength of light :Wavelength of monochromatic light can be calculated by using the formula λ=
where x= band width.
d = distance between two slits.D = distance between slit & screen
I. To find –D. D can be obtained by measuring the distance between slit & eyepieces. This
distance can be measure from scale marked on the optical bench.
II.
To find-X
Band width is measure with the help of micrometre eyepiece. Initially eyepiece is adjusted in 1such a way
that vertical cross wire co-indices with 1 end of bright band & corresponding micrometre reading is noted.
Then rotate micrometre screw so that vertical cross wire is on the successive bright band. Take the
corresponding micrometre reading .Let corresponding reading for bright bands is x1, x2, x3, x4,x5.....
Find the values of (x2 – x1), (x3 – x2), (x4 – x3).....
Average of these values is band width x.
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To measure d convex lens is kept between biprism & eyepiece then move the lens towards eyepiece, such
that two diminished images of slit are clearly seen vertical cross wire of eyepiece is made to co-inside with
each image & corresponding micrometre reading is taken .Let,d1 be the distance between two diminished
images.
Move the lens towards biprism such that two magnified images are seen through eyepiece .Let d 2 be the
distance two magnified images.
From fig(I)
∴
=
=
.
---------1
From fig II
=
∴
= ---------2
Multiplying equation 1 by 2
×
= ×
.
=1
= d1 . d2
∴
d=
.
∴ Wavelength of monochromatic source of light can be calculated by using the formula.
λ=
.
Q.13. what is diffraction. Explain the types of diffraction
Ans:- The bending of waves around edges of obstacles is called as diffraction
I. Fresnel diffraction
II. Fraunhofer diffraction
I)
Fresnel diffraction:-In this type of diffraction sources & screen are at finite distances from
obstacles the incident wave front is not a plane wave front.
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Frounhofer diffraction: - In this type of diffraction sources & screen are at infinite distances
from obstacles. The incident wave front is a plane wave front.
Q.14.Explain qualitatively the intensity variation in diffraction pattern produce by a single slit.
Ans:- Consider, monochromatic Parallel, beam of light of wavelengthFalling on narrow slit of width a. A
diffraction pattern is form on Screen placed in focal plane of convex Lens. Let O be the centre of slit & Po
be thecentral point on screen.
The intensity is to be find at pt P on screen .The maximum intensity is observed at central point Po.
It is called as principle maximum.
As angle θ increases, the intensity changes between alternate minima & maxima .The condition for
minima is
asinθ = nλ , where n=±1,±2,±3,...
Condition for maxima is
a sinθ=
+
λ , where n=±1,±2,±3,...
Q.15 what is Rayleigh Criterion?
Ans:-According to Rayleigh, the images of two pt objects close to each other are regarded as
resolved(separate) if central maximum of one falls on 1st minimum of the other. This is called as Rayleigh
Criterion for resolution.
Diagram:-
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YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
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Physics-II
Q.16 :- Define resolving power of a telescope.
Ans :- The resolving power of a telescope is define as a reciprocal of least angle subtended at the objective
by two distance point objects which are just resolved.
Resolving power (R.P) = = . Where D =Diameter of objective
λ=Wavelength of light.
Q.17.Define resolving power of microscopic.
Ans:- The minimum distance by which the two objects are close to each other so that their images are
separate is called as limit of resolution of a microscope.
The reciprocal of limit of resolution of microscope is called as its resolving power
.
R.P.of microscope = =
Where n =refractive index between object & objective
λ =wavelength of light
α=angle subtended by object with objective.
nsin α =Numerical Aperture(N.A)
Q.18 Distinguish between Interference & Diffraction
Ans:Interference
Diffraction
i)
It is result of superposition of
i) It is the result of limited portion of light
different waves
wave of same wavefront.
ii)
Interference fringes are of same
ii) Diffraction fringes are not of same width.
width
iii)
All bright bands are of equal
iii) The intensity of central maximum is
intensity
highest and secondary maxima are of
decreasing order.
iv)
The region of minimum intensity iv)The region of minimum intensities are
is dark.
not perfectly dark
 R=
+ +2a1.a2 cosθ
Intensity α (amp)2
 IR = + +2a1.a2 cosθ
For constructive interference,
θ = 0, 2π, 4π,...
∴ cosθ =1
Imax =(a1+a2)2
For destructive interference,
θ = π,3π,5π,....
cosθ = -1
Imin = (a1 – a2)2
(
)
=(
Let,
)
=r =amplitude ratio
(
∴
=(
Page | 19
)
)
1
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IR=I1 + I2 +2 . cosθ
If I1 = I2 =I
IR=2I +2√ cosθ
=2I +2I cosθ
=2I(1+ cosθ )
=2I (2cos2 )
IR =4I cos2
Imax =4I
Imin =0
 Important Formulae
.
1)λ =
or
λ=
2)Path difference =nλ .....(for bright point)
Path difference = −
..... (For dark point)
3) R.P.telescope =
.
4)R.P.of microscope =
(
5)
)
=(
6)
)
(
=(
where n sin α =N.A.
*
=
=
)
)
IR=I1 + I2 +2
.
cosθ
Q. Numericals:1) the optical path different between two indentical waves arriving at a point is 80.5 wavelength is the pt
bright or dark if path difference is 4.8 ×10-5m.Find wavelength of light used.
Ans:- Path difference =80.5λ
Path difference=4.8×10-5 m
I)Point is bright or dark
II)λ
I)Path difference =80.5λ
= 80 +
=
=161 ×
Since path difference is odd multiple of half the wavelength .Hence,pt is a dark point.
∴ 80.5 λ=4.8 × 10-5 m
λ=
. ×
.
λ=
× 10
let x=
taking logs log x =log 48 –log 805
Page | 20
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Prof.Anarse.D A
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= 1.6812 -2.9058
=2.7754
X= 5.962 × 10
λ =5.962 × 10 × 10
=5.962× 10
=5962× 10 m
λ =5962 A0
Q. Path difference between two identical waves arriving at a pt is 100.5 λ is the point dark or
bright. if path difference is 44 m.Calculate wavelength.
Ans :- Path difference =100.5 λ
Path difference = 44 m
= 44 ×10-6 m
To find i) Point is bright or dark
ii)
λ =?
i)
Path difference =100.5λ
= 100 +
λ
=
=
=
λ
=
= 201 ×
Since path difference is odd multiple of half the wave length Hence,point is dark point.
And
100.5 λ =44×10-6m
λ=
λ=
×
.
×10-6
put x=
∴ log x =log 440 –log 1005
= 2.6435 – 3.0021
=1. 6414
λ =4379 × 10-1
λ =4.379 × 10-1 × 10-6
= 4.379 × 10-7
= 4379× 10-10 m
∴ λ =4379 A0U
3) Monochromatic light from a narrow slit eliminate two narrow slit 0.3 mm apart producing an
interference pattern with bright fringes 1.5mm apart on a screen 75cm way .find the wavelength of
a light how will the fringe width alter if a)The distance of the screen is doubled
b) The separation between the slits is doubled
Ans:- d=0.3 mm=0.3×10-3m
X= 1.5mm =1.5×10-3m
D =75 cm =75×10-2m
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YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
λ =?
D ‘=2D
d’’=2d
a)
b)
Yashasvi Science Academy
Physics-II
x’=?
x’’=?
1)x =
∴
=
=
=
. ×
×
. ×
×
× ×
×
∴
λ = 0.6 × 10 m
λ = 6000 × 10 m
∴ λ =6000 A.U.
2) x = XαD
∴ =
∴
.
=
∴ x’ =3mm
3) x α
∴ = .
= .
∴ x’’ = ′′ =0.75mm
4) In Young’s double slit experiment the slits are 0.5mm apart & interference is observed on a
screen placed at a distance of 100cm from the slits .It is found that 9th bright fringe is at a
distance 8.835 mm from the 2nd dark fringe the centre of the fringe patern .Find the wavelength
of light used.
Ans:- d=0.5mm=0.5×10 m
D=100cm =1m
X9 –X2 =8.835mm=8.835×10 m
For nth bright band , Xn =n.
For 9th bright band , Xg =
−
For 2nd dark band X2 = 2 −
For nth dark band Xn =
∴ X2 = Page | 22
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- X9 –X2 =
X9 –X2 =
∴ 8.835×10 = ×
×
. ×
8.835×10 =
.
×
∴ λ =
λ = 0.589×10 m
=5890×10 m
=5890A.U
5) A pt is situated at 6.5 cm& 6.65cm from two coherent sources .find the nature of elimination at
the pt if the wavelength of light is 5000A.U.
Ans:- λ =5000A.u.
=5×10 m
Path diff = 6.65 -6.50
= 0.15 cm
=0.15 ×10 m
Path diff = nλ
n=
=
.
×
×
×
= ×
=3 × 10
∴ n =3000
Since path diff is integral multiple of wavelength hence pt is a bright point.
6) In biprism experiment the eyepiece is placed at distance of 1.2cm from the sources the distance
between virtual sources was found to be 7.5×
m .find the wavelength of light if the eyepiece
is to be moved transversally through a distance of 1.888cm for 20 fringes.
Ans:- D=1.2m
d=7.5× 10 m
distance between 20 fringes =1.888cm
λ =?
1)Distance between 20 fringes =1.888cm
.
X=
.
X=
=0.0944cm
X= 0.0944× 10
We know, x=
λ=
=
=
Page | 23
.
×
×
× . ×
.
× 10
=5900× 10 m
∴ λ= 5900A.U
YASHASVI SCIENCE ACADEMY
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7)
Yashasvi Science Academy
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A biprism is palced 5cm from slit eluminited by sodium light of wavelength 5890A.U. The width
of the fringes obtain on a screen 75cm from the biprism is 9.424×
cm what is the distance
between two coherent sources.
Ans:- u=5cm
λ=5890A.U. =5890× 10
v=75cm
x=9.424× 10 cm = 9.424× 10 m
d=?
I)
D = u+v
= 5+75 =80cm
D=0.8m
II)
x=
∴ d=
d=
d=
=
×
.
×
× .
×
×
×
×
× 10
d=5× 10 m
8)
What is the minimum angular separation between two stars if telescope is used to observe them
with an objective of aperture 20cm .The wavelength of light used is 5900 A.U?
Ans:- α =?
D= 20cm =0.2m
λ=5900A.U =5.9× 10
R.P of telescope = = . .
α=
.
=
× . ×
.
×
.
=
× 10
= 0.61 × 59 × 10
α = 35.99× 10 rad
9)
Young’s
Two slits in Ath experiments ,have widths in the ratio 81:1 what is the ratio of the amplitude of
light waves coming from them?
Ans:- W1: W2 = 81:1
a1:a2 =?
We know,
=
∴
=
=
=
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YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
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=
∴ a1:a2 = 9:1
10)
Two coherent sources whose intensity ratio 81:1 produce interference fringes .Calculate the
ratio intensity of maxima & minima in the fringe system.
Ans:- I1:I2 =81:1
=?
∴ Intensity α2 (ampl)2
=
=
∴
=
By using componendo-dividendo
∴
=
=
= ( )
=(
)
=
11) Find the ratio intensities at two pts x & y on a screen in Young’s double slit experimental where
waves from S1 & S2 have path difference at o &
Ans:- I) o II)
I)
Ix =I1+I2+2
. cos θ
Let I1=I2=I, phase diff=0
Ix =I+I+2 . cos θ
=2I+2I(1)
Ix =4I
When Path diff is
path diff =
Iy =I+I+2 .
=2I+2I(0)
Iy =2I
cos.
12) Diffraction pattern of a single slit of width 0.5cm is form by a lens of focal length
40cm.Calculate the distance between 1st dark & next bright fringe from the axis the wavelength of
light used is 4890A0
Ans:- a=0.5cm =0.5×10-2 m
D=f =40cm =0.4m
λ =4890A0 =4890×10-2 m
x2 –x1 =?
For the minima,
a sinθ =λ
For the maxima,
a sinθ= + λ
For the 1st bright & dark fringes,
=λ
Page | 25
YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
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λ
=
=
==
λ–λ
(x2 –x1) =
X2 X1 =
×
×
.
= × . ×
= 4890×0.4×10-2
= 1956 × 10-8 m
13) The semivertical angle of the cone of the rays incident on the objective of a microscope is 200
if the wavelength of incident light is 6600A0.Calculate the smallest distance between two points
which can be just resolved.
Ans :- α =200 λ =6600A0 =6.6×10-7 m
d =?
1) For illuminated light
R.P. of microscope = =
∴ d=
. ×
d=
× ×
. ×
=
d=
.
.
.
× 10
.
Let x= .
log x =log 3.3 +log 0.3420
= 0.5185 +1.5340
=5
d= 9.649 10 m
2)For self luminous light
R.P. of microscope = = . d=
.
= 1.22 = 1.22× 9.649 × 10-7
Page | 26
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12. Electrostatics
Q.1 State & prove Gauss’s Theorem.
Ans:-Diagram
Statement: - For a close surface of any shape with any number of charges situated in any position inside it
the T.N.E.I. (Total normal electric induction) over a close surface is equal to the algebraic sum of electric
charges enclosed by that surface.
Consider a charge +q situated at a pt O inside a conductor of any shape considers small area ds on
surface of conductor at distance r from pt O.
N.E.I over area ds = ∈Ecos θ ds
Ele. Intensity at a dist. r from charge q is E= ∈
N.E.I. over area ds = ∈ =
Let dω =
cosθ ds
∈
= solid angle subtended at charge q by area ds.
∴ N.E.I. over area ds=
∴ T.N.EI = ∫
dω
∴ T.N.EI = ∫ dω
(dω )
∫ dω = 4 = total solid angle subtended a pt o due to closed surface.
∴ T.N.EI = × 4
T.N.E.I = q
If q1,q2,q3, ...... qn be the charges enclosed inside the closed surface then,
T.N.E.I = q1+q2+q3+ ...... +an
T.N.E.I =∑ qi
Hence, Gauss’s theorem is proved.
Q.2. Obtain an expression for electric intensity at a point outside the charge sphere.
Ans:- Consider, a charge +q is given to a conducting sphere of radius R .Charge is uniformly distributed
over the surface of sphere.Consider,any point outside the sphere at a distance r from centre of sphere.Draw
a concentric sphere of radius r electric intensity is perpendicular to sphere for any pt on sphere.
Page | 27
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T.N.E.I for sphere B= ∫ ∈ E cosθ ds
For any point on sphere ,θ =0
Cos 0=1
T.N.E.I for sphere B = ∫ ∈ Eds
T.N.E.I. for sphere B =∈ E ∫ ds
∫ ds =total surface area of sphere = 4πr2
T.N.E.I. for sphere B= ∈ E× 4πr2…………..1
By Gauss’s theorem,
T.N.E.I. = q ………………2
From 1 &2,
∈ E × 4πr2 = q
∴ E= ∈ Let ∈ = ∈ K
Where, ∈ = Permittivity of free space or vaccum
K = dielectric constant
E = ∈ For air medium,
∴ E= ∈ K=1
We know, surface charge density =
∴
=
q=
× 4πR
E=
=
∈
×
∈
E= ∈ For sphere A, r=R
E = ∈
Q.3 Obtain an expression for electric intensity at a point outside the charge cylinder.
Ans:- Consider ,a charge cylinder of radius R having a charge +q per unit length. Consider any point
outside the cylinder at a distance at a distance r from centre of cylinder. Draw an imaginary cylinder of
radius r. Let L be the length of imaginary cylinder. Let E be the electric intensity perpendicular to the
imaginary cylinder.
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No tubes of induction are form at plane surface of imaginary Cylinder. Therefore,TNEI at plane surface of
cylinder is zero.
T.N.E.I = for curved = ∈ × curved surface area of cylinder Surface of area
T.N.E.I = ∈ × 2πrl ………………1
Total charges enclosed by the cylinder = ql
By Gauss’s Theorem
TNEI =ql………………….2
From 1 & 2
∈ × 2πrl = ql
E = ∈× E = ∈× ∴ ∈ =∈ K
Where,
∈ =Permittivity of free space or vacuum
K =dielectric constant
E= ∈ For air med,
K =1
E = ∈ Case i)
Case ii)
Ele.int on the surface of cylinder,
E = ∈ ( r =R)
Ele.int inside the cylinder
E =0
( q=0)
Q.4 Obtain an expression for electric intensity at a point near a charge conductor.
Ans:- Consider, a charge conductor of any shape electric intensity outside the conductor is ⊥ to
conductor while electric intensity inside the conductor is zero.
Consider, a close surface in the form of cylinder having cross sectional area ds.T.N.E.I. Over closed
cylinder passing through its end outside the conductor is
∈ Eds.
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YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
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∴ T.N.E.I. = ∈ Eds ------------ 1)
Let be the surface charge density
∴ (Charge per unit area)
∴ Total charges enclosed by the conductor =
By Gauss’s Theorem,
T.N.E.I. =
ds -------------- 2)
From 1 & 2
∈ Eds = ds
E =∈
Physics-II
ds
E = ∈
∴ ∈ =∈ K
Where,
∈ =Permittivity of free space or vaccum
K =dielectric constant
E = ∈ For air med, K=1
E = ∈ Q.5 Obtain an expression for mechanical force per unit area of charge conductor.
Ans:- Consider a charge conductor of any shape situated in a medium of dielectric const. K. Let ds be the
small area of surface of conductor be the surface charge density.
Electric intensity at a point just outside the conductor is given by
E =
-------------------- 1
Let E1 be the electric intensity due to the charge on small area ds & E2 be the electric intensity due to
charge on rest of the conductor.
E = E1 + E2 = Resultant intensity acting inside the conductor is given by E1 - E2 ,but intensity inside the conductor is 0.
∴ E1 - E2 =0
E1 = E2
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YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
E1 + E2 =
Yashasvi Science Academy
Physics-II
E1 = = E2
Force acting on area ds due to charged conductor.
F = (charge on area ds) × (Ele.int.due to rest of conductor)
F = ds × F=
Mechanical force per unit area,
f=
=
×
∴ f= From equation 1
= E
f= ∴ f=
Q.6 Obtain an expression for energy density in an electric field .
Ans:- When a charge is given to a conductor electric field is produced while charging conductor work is
done in order to bring the charge to conductor this work is stored in the form of electrostatic energy .
We know, mechanical force per unit area is given by
f= force acting on area ds = F = × ds
Let dx be the displacement of area ds against the force
∴ Work = force × displacement
dω = F× dx
dω == .ds × d x
ds × dx = dv =volume traced by area ds.
dω = .dv
But work done can be stored in the form of energy
dE = × dv
=
=
= energy density
∴ Energy density =
Page | 31
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Prof.Anarse.D A
U= Yashasvi Science Academy
Physics-II
Q.7 what do you mean by capacity of a conductor? State the units of Capacitance. Define 1 farad.
Ans:-The ability of a conductor to hold the charge is called as capacity or capacitance of conductor. When
conductor gains an electric charge its potential increase .the charge on conductor is directly proportional to
the potential difference. Let Q be the charge on conductor & v be the potential difference then.
Q∝V
Q =CV
∴C=
Capacity of conductor © is define as quantity of charge required to raise its potential by 1 unit
Unit of C =
∴ Farad =
1F =

1 farad =If a charge of 1 coulomb increase the potential of conductor by 1 volt ,the capacity of
conductor is said to be 1 farad.
Smaller units of capacitance:1 pf = 1 picofarad = 10-12 f
1 nf = 1 nanofarad = 10-9 f
1 f = 1 microfarad = 10-6 f
Q.8 Explain the types of condenser.
Ans:- There are 3 types of condenser
Types of Condenser:1) Parallel Plate Condenser
2) Spherical Condenser
3) Cylindrical Condenser
1) Parallel Plate Condenser :It consists of two parallel plates separated from each other by air or any dielectric material. The
charge is given to one plate while other plate is grounded.
2) Spherical Condenser:It consists of two concentric spherical conductors separated from each other by air or any dielectric
material. The charge is given to inner sphere & outer sphere is grounded.
3) Cylindrical Condenser:Page | 32
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It consists of two co-axial cylindrical conductors separated from each other by air or any dielectric
medium. The charge is given to inner cylindrical & outer cylinder is grounded.
Q.9 Explain the concept of condenser.
Ans:A
B
A
+
- +
+
- +
+
- +
+
- +
+
- +
B
+
+
+
+
+
-
Capacity of charge conductor increase if another conductor connected to earth is kept near it; this is the
principle of condenser.
Consider a parallel plate condenser consist of two parallel metal plates separated from each other. It
positing charge q is given to plate A, then equivalent negative charge is induced on near side of plate B and
+q charge is induce on free charge +q moves towards earth. Therefore, only negative charge is left on plate
B.Thus potential of plate A decreases & hence its capacity increases.
Let C be the initial capacity of plate A.Q be the charge & V be the potential difference.
C = -------------1
But, when negative charges are induce on plate B negative potential
-V1 is produced on plate A
C1 =
-----------2
From equation 1 & 2
C1 > C
Hence, capacity of conductor increases.
Q.10 Derive an expression for capacity of parallel plate condenser filled with dielectric K.
Ans:- Parallel plate condenser consist of two plates A & B separated by distance d.Let K be the dielectric
constant plate A is given +Q charge .Therefore, -Q charge is induced on inner surface of plate B. Plate B is
grounded. Electric intensity is perpendicular to plate A.
+Q
A
+ +++++++++++++++
E
- -- - - - - - - - - - - - - - - - - - - _________________________
B
Page | 33
YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
Yashasvi Science Academy
Physics-II
By using Gauss’s Theorem,
E = Where,
= Surface change density
=
E = V = (Electric Intensity) × (dist between two plates)
V =E × d
V=
×d
C =
C=
C=
Q.11. Derive an expression for energy stored in charge capacitor.
Ans:- When capacitor is charged some work is required for charging the capacitor. This work is stored in
the form of electrostatic energy. Let Q be the charge & V be the P.D.
Q =CV
V=
Let q be the charge & v be the P.D. at any time t,
∴v=
Let dω be the work done in displacing additional charge dq.
∴ dω =v × dq
∴ dω = × dq
Total work done can be finding out by integrating this relation within the limit 0 to Q.
∫
=∫
dq
∫
W=
.dq
W=
− =
W= ×
But, work done can be stored in the form of energy.
E=
Q = CV
E=
E=
CV2
C =
E = × V2
E = QV
Page | 34
YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
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Physics-II
Q.12. Derive an expression for equivalent capacitors when three condensers are connected in
parallel.
Ans:- In parallel arrangement one terminal of each condenser is connected to positive terminal of battery &
other terminal is connected to negative terminal of battery.
Let Q1, Q2, Q3, be the charges C1, C2, C3,resp
Let V be the P.D. across each condenser.
From the conservation of charge,
Q = Q1+ Q2 +Q3
Q =CV
Q1 = C1V1 Q2 = C2V1 Q3 = C3V1 Q4 = C4V1
CV = C1V + C2V +C3V +C4V
CV =V (C1 +C2+ C3 )
C = C1 +C2+ C3
In general if n capacitors are connected in parallel
Cp = C1 +C2+ C3+ -------+Cn
Thus, in parallel arrangement the equivalent capacitors are equal to sum of their individual capacitance.
Q.13.Derive an expression for effective capacitors or 3 capacitors connected in series.
Ans:- If the capacitors are connected one after another then the arrangement is called as series
arrangement Let C1, C2 & C3 be the capacities of 3 condensers. Positive plate of 1st capacitor is connected to
positive terminal of battery and negative plate of last condenser is connected to negative terminal of
battery. Let V1, V2 &V3 be the potential difference across capacitors of capacitance C1, C2 & C3 resp.Let V
be the potential difference across series combination.
From the fig,
V = V1+V2 +V3
Q =CV
V =
V1 =
=
, V2 =
+
Page | 35
, V3 =
+
YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
=
Yashasvi Science Academy
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+ + = + +
In general, if n capacitors are connected in series.
= + + + ------+
This series arrangement reciprocal of equivalent capacitors is equal to sum of reciprocal of their individual
capacitances.
Q.14..with the help of neat diagram. Describe the construction, working and uses of the Can de
Graff generator.
Ans:- Diagram
P1, P2 =Pulley
B1B =Conveyer belt
A =Spray brush
C =Collector brush
D =Hollow dome shaped conductor
E =Evacuated accelerating tube
I =Ion source
P =DC power supply
S =Steel vessel filled with N2
M = Earthed metal plate.
If a charge is continuously supply to an insulted metallic conductor, the potential of conductor goes
on increasing; it is used to developed very high potential of the order of 10-7 volt
 Construction :Conveyer belt (B1B) made of an insulting material like a rubber or silk can move 2 pulleys P1&P2.The
belt is driven by an electric motor connected to the lower pulley P1.
The spray brush A consists of large no of pointed wires. It is connected to positive terminal of high
voltage Dc supply. From this brush positive charge can be spread on belt. This charge can be collected
by another similar brush C. This brush is connected to a large dome shape hollow metallic conductor
D which is mounted on insulating pillars.
E is an evacuated accelerating tube having an electrode I at its upper end. This electrode is connected
to dome shaped conductor. To prevent lickage of charge from the dome, it is enclosed inside a large
steal vessel S.This vessel is filled with nitrogen at high pressure .A Freon gas is mixed with nitrogen to
have better insulation between the vessel S and its contents metal plate M held opposite to brush A on
the other side of belt is connected to vessel S which is grounded.
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YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A

Yashasvi Science Academy
Physics-II
Working:The electric motor connected to the Pulley P1 is switch on the pulley begins to rotate and the conveyer
belt is set into motion .The Dc power supply is then switched on from the spray brush A positive
charge is continuously spread on belt B. The belt carries this charge in upward direction. The charge
collector brush C collects this charge and send it to the Dom shape conductor. The charge is
distributed over the outer surface of dome.because of continuous accumulation of the charge its
potential increase to high value. The potential of electrode I also rises to high value. The positive ions
such as protons or direction from a small vessel containing ionised hydrogen are introduced in upper
part of evacuated accelerating tube. These ions repelled by the electrode I are accelerated in downward
direction due to the very high fall of potential .These ions acquire very high energy.

Uses: - Van de Graff generated is used to produce high energy charge particles of the order of 10 mev.
They are used
I) To carry out disintegration of nuclei.
II) To produce radioactive isotopes.
III) To study nuclear structure
To study different types of nuclear reactions.
Q.15) Explain the effects of dielectric on capacity of a parallel plate capacitor.
Ans:- Consider,a parallel plate capacitor of area A
Let d be the distance between two plates and dielectric medium air between them
Cair =
------------------ 1
If dielectric material of dielectric constant K is introduced between two plates then,
Cd =
Divide equation 2/1
=
× =K
∴ Cd =K.Cair
When dielectric is inserted in the space between parallel plate of a charge capacitor, surface charge on
conducting plate does not change but induce charge of opposite side appears on each surface of dielectric.
Let be the surface charge density on plates of capacitor & I be the surface charge density of
induced charged. therefore net surface charge density on each side of capacitor has a magnitude.
= −
Electric field between the plates of capacitor without dielectric,
E0 =
Electric field between the plates of capacitor with dielectric with dielectric is
E=
= ---------- 3
Also, E =
=
----------4
From equation 3 & 4
=
=
=
=
= Page | 37
(K-1)
(
)
YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
Yashasvi Science Academy
Physics-II
1−
=
Q.16) what do you mean by polar & non polar molecules?
Ans: - Polar molecule –i) in a polar molecule, centre of gravity of positive nuclei & revolving electrons do
not coincide.
i)
Polar molecules have a permanent electric dipole movement eg. HCl,H2O,N2O
IV)
5) Metal sphere of radius 20cm is charged with 12.56 c situated in air. Find the surface density of
charge. Calculate the distance of point from centre of sphere. Where electric intensity is 1.3×105
N/C.
Ans:- R=20cm =2×10-1
q= 12.56 =12.56×10-6C
E=1.13×10-5N/C
K=?
1) =? 2) r=?
I) = =
.
=
=
.
×
× . × ×
×
.
×
× 10
=
= 0.25× 10 c/m2
ii) E =
r2 =
r=
=
=
=
×
.
×
.
× .
.
× .
× .
×
× × .
×
×
= 1.0000 -0.9469+0.0531
= 1.0000-1.0000
= 1.000
=√1 = 1
Q.6) Capacitor consist of two parallel metals plates each at areas 100cm2 separated by a dist
2mm.what is capacitance of capacitor if charge to 1000µm.What will be P.D. between two plates.
how will the capacity of air capacitor be affected if the space between plates is completely filled
Mylar.
Kmylar =3
Ans:-A =100 cm2 =100× 10 =× 10 m2
d= 2mm= 2 × 10 m
Q=1000 µm =10 C. Kmylar =3
Page | 38
YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
Yashasvi Science Academy
Physics-II
i)C=? ii)V=? iii) Cmylar=?
i)
C=
× .
×
=
×
∴ C =4.425 × 10
ii)
×
Q =CV
V=
=
.
=
×
.
= 0.2259 × 10 V
iii)
C∝ K
=
.
=
=
×
=
Cmylar = 4.425 × 3 × 10 = 13.275 × 10 Q.7) The energy density at a point in a dielectric constant 8 is 26.55 ×
field intensity at that point.
Ans:- K=8
U= 26.55× 10 J/m3
E=?
U
=
=
E=
=
×
.
.
×
.
=
Let x =
Log x = [
×
×
.
.
=
E
×
.
× 10
.
=
.
× 10
.
.
26.55 −
35.40]
= [1.4240 − 1.5490]
Page | 39
YASHASVI SCIENCE ACADEMY
J/m3.Calculate electric
Prof.Anarse.D A
Yashasvi Science Academy
Physics-II
= × 1.8750
.
=
=
+
.
= 1+0.9375
Log x = 1.9375
x = 8.660 × 10
x = 0.8660
E = 0.8660 × 10 N/C
Q.8) A metal plate of area 0.01m2 carries a charge of 100µc.Calculate the outward pull of plate.
Ans:- A=0.01m2 = × 10 m2
Q = 100 µc = 10 C
Outward pull = force =?
F=
= , ds = A
×
F=
F=
F=
× .
=
=
×
×
×
.
.
×
= .
F = 0.5650× 10 N
F = 56500 N
Q.9) Circular metal plate of radius 1.5 cm is charged with 10µc situated in air. Calculate mechanical
force acting on it if the same circular plate carrying same charge were kept in Acetone what should
be the value of mechanical.
Ans:- R =1.5cm =1.5 × 10 m
Q =10 µc = 10 c
Kacetone=27
i)F=? 2) Facetone=?
ii)
F=
= , ds = A
F=
×
F=
A =2 R2
Page | 40
YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
F=
(
=
=
=
=
Yashasvi Science Academy
Physics-II
)
× .
× .
× .
.
×
× .
×
×
× .
× .
= log1-[
8.85]
28.278 +
= 0.0000 - [1.4513 + 0.9469]
= 0.0000 – 2.3982
=7.6018
= 3996 N
F
2) F∝
=
=
=
= 148 m
Q.10) Long cylinder of radius 2m carries a charge of 5µc/m kept in a medium of dielectric constant.
Find the electric field intensity at a point situated at a distance 1 m from the axis of cylinder.
Ans:- R =2m
q= 5µc/m = 5× 10 c/m
r=1m
K=1,
E =?
E =
=
×
× .
× .
×
×
×
106
E
= . × . ×
Log E = log 5 – (log 6.284 +log8.85)
E = 0.08991×106
= 89.91× 103
Q.11) A thin long cylinder of radius 1 cm carrying a charge 5µc/m kept in water. Find the electric
intensity at a point situated at a distance 10cm from the axis of cylinder Kwater is 81.
Ans:- R =1cm =0.01cm
Page | 41
YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
Yashasvi Science Academy
Physics-II
5×10-6
q = 5µc/m =
c/m
E =? r=10 cm=0.1m
K= 81
E=
=
Q.12)A network of 4 capacitors 5µc/m each are connected to a 240V supply Determine
i) Equivalent capacitance of network
ii) The chatge on each capacitor
Ans:- I) C1, C2 and C3 are connected C1 in series.
= + +
= + +
∴ Cs =
=
∴ Cs =1.666 µ
ii)Cs and C4 are connected in parallel
Cp = Cs + C4
Cp = 1.666+5
∴ Cp=6.666 µ
III) In series arrangement charge on each capacitor is same
Q= CsV
=1.666× 10-6 × 240
= 399.84× 10-6 c
IV) Charge on C4
Q= C4V
= 5× 10-6×240
= 1200× 10-6C
Q.13) 3 Capacitors are 10 µ ,15 µ ,20 µ are connected in series & then in parallel with a 250V
supply. Calculate i) Equivalent Capacitance in each case
ii) Charge on each condenser in each case.
Ans:- C1=10 µ , C2=10 µ , C3=10 µ , V=250V
I)For series arrangement,
= + +
=
+
+
=
=
Cs =
∴ Cs = 4.615 µ
II) For parallel arrangement,
Cp = C1+C2 +C3
= 10+15+20
Cp = 45 µ
iii) For series arrangement charge on each condenser is same
Page | 42
YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
Yashasvi Science Academy
Q = CsV
= 4.615 × 10-6×250
= 1153.75× 10-6c
iv)
For parallel arrangement charge on each condenser is different
i)Q1 =C1V
= 10× 10-6 × 250
= 2500× 10-6c
ii) Q2 =C2V
= 15× 10-6 × 250
= 3750× 10-6c
iii) Q3 =C3V
= 20× 10-6 × 250
= 5000× 10-6c
Page | 43
YASHASVI SCIENCE ACADEMY
Physics-II
Prof.Anarse.D A
Yashasvi Science Academy
Physics-II
13. Current Electricity
Q.1 State and explain Kirchhoff’s current law. (1st law or Junction law)
Ans: - Statement: - The algebraic sum of current at any junction is equal to zero. Mathematically, ∑ = 0
Current approaching towards the junction is treated as positive and current leaving the junction is treated
as negative.In the figure, I1 and I2 are currents approaching towards junction and hence positive where, I3
,I4 ,I5 are leaving the junction and hence negative,By applying Kirchhoff’s law,
I1 +I2 -I3- I4 -I5 = 0
∑ I =0
Q.2 State and explain Kirchhoff’s Voltage law (2nd law/loop theorem)
Page | 44
YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
Yashasvi Science Academy
Physics-II
Ans:- Statement :- The algebraic sum of potential differences(product of current and resistance)and
electromotive force applied in a close loop of electrical network is zero.
Mathematically,
∑ IR + ∑ E =0
If the direction of tracing is same as that of conventional current flow, the P.D.across the resistance is
considered as negative otherwise it is positive.
E.M.F. is positive if we travel from negative terminal to the positive terminal inside the cell and E.M.F. is
negative is we travels from positive terminal to the negative terminal.
Consider, a close loop AFEDCBA,
-I1R1+I2R4 –E2+I2R5-I1R3+E1 = 0
I2(R4+R5)- I1(R1+R5)+ (E1+E2) =0
Consider, a close loop AFCBA
-I1R1 – (I1 + I2)R2 - I1R3 + E1 =0
-I1(R1 + R2+ R3) - I1R2 + E1 =0
Consider, a closed loop of FEDCF
I2R4 – E2 + I2R5 +(I1 +I2)R2 =0
I2 (R4 +R5 +R2 )+ I1R2 – E2 =0
Q.3 With the help of neat diagram obtains the balancing condition of Whetstone’s network.
Ans: - simple circuit given by Wheatstone to find unknown resistance is called as Whetstones network. It
consist of 4 resistances, R1 ,R2 ,R3 &R4 to form a close loop,ABCD battery E is connected between
diametrically opposite junction A &C
Page | 45
YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
Yashasvi Science Academy
Physics-II
Galvanometer is connected between pt B & D since,
Potential at B is equal to potential at D.
Therefore, Galvanameter shows null diffraction. The current I supplied by the cell is divided into I1 & I2 at
pt A1 current I1 flows through resistance R1 & current I2 flows through resistance R3 if network is balanced
no current flows through B & D therefore, same current I1 flows through resistance R2 and current I2 flows
through resistance R4.Let VA,VB,Vc & VD be the potentials at point A,B,C,& D resp.
(Potential at B) = (Potential at D)
VB =VD
VA-VB = VA - VD
VB-Vc = VD - VC
From the fig,
VA-VB = I1R1
VA-VD = I2R3
VB-VC = I1R2
VD-VC = I2R4
∴ I1R1 = I2R3 ---------- 1
I1R2 = I2R4 ---------------- 2
Divide equation ½
=
=
Q.4 Explain the use of Whetstones Meter Bridge to determine an unknown resistance.
Ans: -
Page | 46
YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
Yashasvi Science Academy
Physics-II
Meter Bridge is a simple experiment to determine unknown resistance it consists of uniform conducting
wire having 1 m in length; it is fitted on wooden board between two thick copper strips. A third copper
strip is fitted as shown in the fig. Unknown resistance X is connected in one gap and known resistance R is
connected in another gap. A cell, key and rheostat at connected in series with wire. One terminal of
galvanometer is connected at point B and another terminal is connected to the sliding key or jock key.
Connect known resistance R in circuit touch the jockey at different points on wire Ac mark the
point at which Galvanometer shows null deflection let the pt be D measure length AD and DC.
=
Let
=
be the resistance per unit length of wire,
X=R
Q.5 Describe Kelvin’s method to determine resistance of galvanometer by using Whetstone’s Meter
Bridge.
Ans:-
Resistance of galvanometer find by galvanometer A galvanometer G whose resistance is to be determined
is connected in 1 gap & known resistance is connected in other gap of Whetstone’s metebridge.Battery key
& Rheostat are connected in series which metallic wire A junction of resistance & galvanometer is
connected to metallic wire by Jockey.Take a suitable known resistance close the key K and take the reading
of deflection in Galvanometer without touching Jockey to metallic wire, then touch the jockey to the
different points of wire & find pt D where galvanometer shows equal deflection .Hence,Klevin’s,method is
called as equal deflection method. Measure length lg & lr Let G be the resistance of galvanometer.
=
Let
Page | 47
be the resistance per unit length of wire,
YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
=
Yashasvi Science Academy
Physics-II
G = R.
Q.6 what are the errors in determining unknown resistance by Whetstone’s meter bridge? How the
errors can be minimizing?
Ans: - 1) Errors: - i) we assume that wire am uniform &have same resistance per unit length, but if wire is
not uniform errors may cause in the determination of unknown resistance.
ii) Metallic wire is fixed to the copper strips, so that contact resistance may produce.
2) Ways to minimize the errors:i) By interchanging position of X & R and then repeating the experiment.
ii) By selecting value of R such that null point is obtain close to the centre of wire.
Q.7 Explain the principle of Potentiometer.
Ans: - Consider, a long uniform wire stretched on wooden board, Let l is the length of wire and R is the
resistance of wire. A cell E is connected across wire Let VAB be the potential difference across wire.
VAB =IR Let
VAB =I L
be the resistance per unit length of wire,
I=
---------- 1
Consider, the potential difference t any pt P such that distance AP =land r be the resistance of AP.
VAp = Ir
VAp =I L
And L are constant
=
× l
=
×l
∝l
Principle of potentiometer: -The potential difference between any two points of a wire is directly
proportional to length of wire between theses points.
Page | 48
YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
Yashasvi Science Academy
Physics-II
Q.8 .How will you compare E.M.F. of two cells by potentiometer by individual cell method?
Ans :-
Potentiometer consist of thin long uniform wire stretched on wooden board, Battery key and revolted are
connected between pt A & B.Revostat is used to control the current of wire E1 & E2 be the E.M.F. of two
cells to be compared positive terminal of E1 is connected to pt A, which is connected to positive terminal of
battery Negative terminal of E1 connected to Galvanometer which is further connected to Jockey when key
is close jockey is moved on the wire so that Galvanometer shows null deflection
Let P be the pt P at which Galvanometer shows null deflection .Let dist AP =l1 Let I be the current
flowing through the circuit & be the resistance per unit length of wire.
E1 = I l1 ---------1
Replace battery E1 by E2 and repeat the same procedure .Let l2 be the distance at which galvanometer
shows null deflection.
E2 = I l2 -------------------- 2
Divide the equation 1 by 2
∴
=
=
Q.9 Explain the sum & difference method to compare E.M.F. of two cells by potentiometer.
Ans: -
Page | 49
YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
Yashasvi Science Academy
Physics-II
Potentiometer consist of the thin long uniform wire stretched on a wooden board, battery key & rheostat
are connected between pt A & B.
Rheostat is used to control the current of wire .Let E1 and E2 be the E.M.F.’s of two cells to be
compared positive terminal of E1 is connected to point A. which is connected to positive terminal of
battery. Negative terminal of E2 is connected to one end of Galvanometer which is connected to Jockey.
When key is closed, move the jockey on wire so that Galvanometer shows null deflection. Let P be
the point at which Galvanometer shows null deflection & d(AP) =l1.Let I be the current flowing through
the circuit & be the resistance per unit length of wire.
∴ E1 + E2 = I l1 ---------1
Then connect the cell as ()
Again obtain null point by moving jockey on wire, Let l2 be the distance at which galvanometer shows null
deflection,
E1 - E2 = I l2 ---------1
Divide equation 1 by 2
=
=
By using componendo-dividendo
=
=
=
Q.10 Explain with the help of neat circuit diagram how an internal resistance of a cell can be
determine by using potentiometer.
Ans: -
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YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
Yashasvi Science Academy
Physics-II
a source of E.M.F. is connected to and a rheostat Rh E.M.F. of source is greater than E.M.F. of cell,
positive terminal of key is connected to pt A which is further connected to positive terminal of s.
Negative terminal of E is connected to galvanometer & galvanometer is connected to jockey .Resistance R
& key K2 are connected in parallel with cell E.
Initially close the key K1 but key K2 open then move the jockey on wire, so that Galvanometer
shows null deflection. Let l1 be the distance at which galvanometer shows null deflection .since current
does not flow through cell E. Hence fall of potential along length l1 is equal to E.M.F. of cell. Then close
the key K2 and again obtain null pt by same procedure. Let l2 be the distance at which galvanometer shows
null deflection. Since, current flows through cell fall of potential along length l2 is equal to potential
differences of cell. Let I be the current flowing through resistance R & r be the internal resistance of cell.
=
--------- 1
By ohm’s law,
I=
I=
------------------2
------------------ 3
From equation 2 & 3
=
By using Alternant,
=
-------------- 4
From equation 1 & 4
=
By using Dividend,
=
=
r= R
r= R
−
r= R
−
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YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
Yashasvi Science Academy
Physics-II
Q.11 What are the advantage of potentiometer over voltmeter?
Ans:- i) By using very long wire fall of potential per unit length can be very small in case of potentiometer
.Thus, potentiometer can be used to measure very small potential difference.
ii) If a voltmeter is connected across the cell a small current flows through the cell .Hence, voltmeter
measures terminal P.D. of cell but in case of potentiometer ,null point is obtained ,Hence current does not
flows through cell, so that E.M.F. of cell can be measured with potentiometer.
iii) Accuracy of voltmeter cannot be increase beyond certain limit, but accuracy of potentiometer can be
increase up to a certain limit.
Q.12 State the precaution which must be taken while performing experiment with potentiometer.
Ans: - i) The E.M.F. of battery must be greater than E.M.F. to be compared i.e.E>E1,
E>E2 , E > E1 +E2
ii) The positive terminal of E1 or E2 or of the combination must be connected to that end of
potentiometer wire where positive terminal of battery is connected.
iii) The potentiometer wire must be uniform.
iv) The resistance of potentiometer wire should be high.
*Important Formulae *
1)
=
(Balancing Condition)
2) X =R.
3) G = R.
4)
=
5)
=
....... (sum & diff method)
6) r= R
r= R
(Individual cell method)
−
7) V =IR
8) I =
Q.13) Solve Numerical
1) A voltmeter has resistance of 100 what will be its reading when it’s connected across a cell of E.M.F. 2
V & internal resistance 20 .
Page | 52
YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
Yashasvi Science Academy
Ans:- R = 100
Physics-II
r =20
E=2V
V =?
V =IR
V=
=
× 100
V =
V = 1.667V
Q.2 .Potentiometer wire has a length of 2 m and resistance of 10 . It is connected in series with resistance
990 and a cell of E.M.F. 2 volt. Calculate the potential gradient.
Ans:- L=2m, R= 10
, R1=990
,E =2V
Potential Gradient = = ?
=
= ×
×
=
=
= 10-2v/m
Q.3 Four resistance 5 ,10 , 15 unknown x are connected in series so as to form Whetstones’
network Determine the unknown resistance x if the network is balance with these numerical values of
resistance.
Ans: - R1=5 , R2=10 , R4=15 , R3=X
∴ Network is obtained
=
∴
=
∴X=
Page | 53
× 15
YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
X=
Yashasvi Science Academy
Physics-II
X= 7.5
Q.4 Two resistance x & y in the two gaps of a meter bridge gives a null pt dividing the wire in the ratio
2:3.If each resistance is increase by 30 .The null pt divide the wire in the ratio 5:6 .Calculate each
resistance.
Ans:- 1)
=
X = -----------1
2)
=
∴
=
(
)
∴
=
=
∴ 12Y +540 = 15Y+450
540-450=15Y+12Y
90=3Y
∴ Y = 30
Sub value in equation 1
X = × 30
X = 20
Q.5 Potentiometer has length of 1.5m & resi 10 .It is connected in series with a cell of E.M.F. 4 V at
internal resistance 5 .Calculate potential drop per unit length of wire.
An: - L=1.5m
R = 10
r=5
=?
E=4V
∴
=
=
×
=
Page | 54
×
.
YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
=
×
=
Yashasvi Science Academy
Physics-II
.
.
Log = log 40 – log 22.5
= 1.6021- 1.3522
=0.2499
= 1.778 v/m
Q.6 Potentiometer wire has length 10m and a resistance 20 and it terminals are connected to a battery
of E.M.F. of 4 & internal resistance 5 .What are the distance at which null points are obtain when
two cells of E.M.F. 1.5V & 1.3V are connected so as to a) assist b) opposite each other
Ans:- L=10 m,
R =20
, E =4 V
E1=1.5v E2 = 1.3V
r=5
E1 =I l1
E1 =
a)For assist condition,
E1+E2 =
1.5+1.3 =
2.8 =
∴
=
. ×
.
∴
=
∴
= 8.75m
b)for oppose condition,
E1-E2 =
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YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
Yashasvi Science Academy
Physics-II
1.5 -1.3 =
0.2 =
∴
=
=
. ×
= 0.625m
Q.7 A cell of E.M.F. 3 v and internal resi 4 is connected to two resi 10
the cureent through each resistance using Kirchoff’s law.
Ans:-Diagram
By using Kirchoff’s voltage law in the closed loop
AFCBA,
-10
-4( +
) +3=0
-10
-4 −
+3=0
-14
-4
14 + 4
= -3
= 3 ----------------------- 1 ×7
By using Kirchoff’s voltage law,
In closed loop AEDBA,
-24
-4( +
)+3=0
-24
-4 −
+3=0
-4
4
- 28
- 28
= -3
= 3 ----------- 2
Multiplying equation 1
98
4
+28
+ 28
∴ 94
= 21
=3
= 18
=
=
=?
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YASHASVI SCIENCE ACADEMY
& 20 join in parallel .Find
Prof.Anarse.D A
Yashasvi Science Academy
Physics-II
= 0.1914A
Sub this value in equation 1
14(0.1914) +4
4
=3
=3-14(0.1914)
=3-2.6796
4
=0.3204
= 0.0801 A
Q.8 current flowing through an external resi of 2 is 0.5A when it is connected to the terminals of
cell.This current reduces to 0.25a.when the external resi is 5 .Use Kirchoff’s laws to find E.M.F. of cell.
Ans:-By using Kirchoff’s voltage,law in the closed loop AFCBA,
-0.5(2) -0.75(r)+E =0
-1-0.75r+E =0--------------------1
1+0.75 r-E =0
By using Kirchoff’s voltage law in closed lop AEDBA,
-0.5(5) -0.75(r)+E =0
-1.25-0.75r+E =0
1.25+0.75 r-E =0---------------2
Subtracting equation 2 from 1
1+0.75 r-E =0
+
1.25+0.75 r-E =0
-0.75
Q.9 4 ressitance of 8 & 12 on a Wheatstone’s network what shont will be needed across 12
to balance the network (shont –parallel)
Ans:- R1= 8 , R2= 8 , R3= 8 R4= 12
Let resistance x is connected in parallel with R4
= +
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YASHASVI SCIENCE ACADEMY
resistance
Prof.Anarse.D A
Yashasvi Science Academy
Physics-II
=
R=
But effective resistance should be 8
8=
8 = (12 + X) = 12X
96 +8X = 12X
96 = 4X
X =24
Q.10) In a meterbridge experiment in R 1 in left gap & resi x in right gap null pt is obtained at 40 cm from
left end with resi R2 in left gap and same resistance X in right gap,null pt is obtained at 50cm from left
end where will be the null pt if R1 & R2 are connected in series in left gap with resi X in right gap.
Ans:- i)
=
R1 = &
ii)
=
R2 = iii)
=
∴
=
=
5(100-l) =l3
500- 5l = 3l
500 = 8l
L = 62.5
Q.11 Two cells having unknown E.M.F. E1 & E2 (E1 >E2) are connected in potentiometer so as to assist
(sum) each other,so that neutral pt is obtain at 8.125m from high potential ,when E2 is connected so as to
oppose E1 neutral pt is obtained at 1.25m from same end compare E.M.F. of two cells.
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YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
Ans: -
= 8.125m
= 1.25m
Yashasvi Science Academy
=?
=
=
=
.
.
.
.
.
=
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.
=
YASHASVI SCIENCE ACADEMY
Physics-II
Prof.Anarse.D A
Yashasvi Science Academy
Physics-II
CH.14. Magnetic Effect of electric Current
Q.1.State ampere’s law.
Ans:- The line integral of magnetic field around any close path is equal to µ0 times the total current passing
through that close path.
Mathematically,
⃗. ⃗ = µ0 I
Where µ0 =Permeability of free space (vacuum)
If θ is angle between ⃗ & ⃗ the line integral can be written as
Bdl cos θ = µ0 I,
Q.2 Using Ampere’s law, obtain an expression for magnetic induction at any pt. near a straight
conductor carrying a current.
Ans:- diagram
Consider, a long straight conductor carrying current I, magnetic induction is to be find at pt.
P which is at a distance r from conductor.
Consider, a circular amperial loop of radius ‘r’ drawn in a place ⊥er to the straight conductor at
every pt. of loop magnetic induction ⃗ is directed in tangential direction.therefore,angle between ⃗ and ⃗
is O at all the points.The line integral along closed loop is given by
⃗. ⃗ = Bdl cos θ
⃗ is along the direction of ⃗.
∴ θ=0,
Cos 0 =1,
⃗. ⃗ = Bdl
= B dl
dl =circumference of loop =2πr
⃗. ⃗ = B × 2πr --------------- 1
By Ampere’s law,
⃗. ⃗ = µ0I --------------------- 2
B × 2πr = µ0I
B=
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Q.3 By Ampere’s law, Derive an expression for magnetic induction along the axis of long straight
solenoid.
Ans:- Diagram
Consider a long straight solenoid. Let n be the number of turns per unit length when current I is passed
through solenoid. Magnetic induction is produced inside the solenoid. Consider a rectangular path ABCD.
Let AB=L be the length of rectangular path .Therefore, number of turns enclosed by rectangle
ABCD =nL.
Therefore, total current flowing through rectangular path =nLI
∴ By Ampere’s law,
⃗. ⃗ = μ nLI ------------------ 1
For close loop ABCDA,
⃗. ⃗ = ∮ ⃗. ⃗ + ∮ ⃗. ⃗ + ∮ ⃗. ⃗ +∮ ⃗. ⃗ The direction of ⃗ ⊥er to BC and AD
∴ ∮ ⃗. ⃗ = ∮ ⃗. ⃗ = 0
Also outside the solenoid magnetic field is very weak(practically O)
∮ ⃗. ⃗ = 0
⃗. ⃗ = ∮ ⃗. ⃗ = ⃗ ∮ ⃗ ∮ ⃗ =L
∮ ⃗ . ⃗ =BL------------------ 2
From equation 1 & 2
μ nLI = BL
I=
B = μ nI
Q.4 Derive an expression for magnetic induction along he axis of Toroid.
Ans:- diagram
Toroid is a solenoid bent into a shape of hollows doughnut .consider a toroid having center O, radius r &
carrying current I. Consider an amperior loop of radius r.current I travels along clockwise direction .Let N
be the total no of turns of toroid. Therefore, total current flowing through toroid = nI
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Physics-II
By Ampere’s law,
⃗. ⃗ = μ nI ------ 1
⃗. ⃗ = Bdl cos θ
∴ ⃗ and ⃗ are in the same direction,
θ =0
cos o=1
⃗. ⃗ = Bdl
=B dl
dl = circumference of loop = 2πr
⃗. ⃗ = B×2πr --------------------- 2
From equation 1 & 2
μ nI = B.2πr
B= .
Let n be the no of turns per unit length
n=
B = μ nI
Q.5 Explain construction & working of suspended type of moving coil Galvanometer (M.C.G.)
Ans:- diagram
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Construction:i) It consist of rectangular coil of thin insulated copper wire suspended between strong horse shoe
magnet .the coil is free to rotate around a core which is fixed between two magnets. The coil is
suspended by a thin wire of phosphor bronze. The current enter the coil through fibre & lives through
helix.(spring)
Mirror (M) is kept between screw & coil to see the rotation of coil by lamp & scale arrangement.
Working:Let rectangular frame PQRS of Length l & breadth ‘b’ is kept in magnetic field of induction B. Let I
be the current flowing through coil. Sides PS & QR are parallel to direction of field.
Hence, they do to experience any force, but sides PQ and SR are perpendicular to direction of magnetic
field. Hence, force acting on them is –
F = BIL
⊥ )×
Torque = (
=BIL × b
∴ = BIA
If coil has ‘n’ turns
∴ = BInA ---------------- 1
∴ Since, coil rotate s it exerts restoring torque on coil .Let θ be the deflection and K be the restoring torque
per unit twist.
Restoring torque = Kθ------------------ 2
At equilibrium,
Deflecting torque = Restoring torque
BInA = Kθ
I=
θ
∴=
is constant
I∝θ
Principles of M.C.G.
Deflection of coil in M.C.G. is directly proportional to current flowing through it.
Q.6 Explain the term sensitivity of M.C.G.
Ans:- Galvanometer is said to be sensitive if it gives large deflection for a small change in colour.
Let dθ be the change in deflection for change of current dI then is called a sensitivity of
galvanometer.
We know,
For M.C.G. I =
Diff w.r.t. I
θ
I=
=
Thus sensitivity of M.C.G. depends upon
i)
Number of turns of coil
ii)
Magnetic induction
iii)
Area of coil
iv)
Restoring torque per unit twist(K)
Sensitivity of MCG can be increased by
a) Increasing number of turns of coil
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b) Increasing magnetic induction
c) Increasing area of coil
d) Decreasing restoring torque per unit twist.
Q.7 Explain the term accuracy of MCG.
Ans:- The relative error in the measurement of current gives accuracy of MCG If relative error is less
accuracy is more.
Suppose any error dθ produces an error dI in the measurement of current I, then is the relative
error in the measurement of current.
We know,
For MCG. I =
θ ------------------ 1
Diff w.r. θ
=
(1)
dI =
dθ ---------------------- 2
divide equation 2/1
=
=
Thus, if deflection θ increases accuracy of MCG increases.
Q.8 Explain how MCG is converted ammeter derive the accuracy formula.
Ans:- diagram
E.M.G. design to measure a current is called as ammeter Galvanometer is converted into ammeter
by connecting low resistance in Parallel with its coil.
Let G be the resistance of galvanometer & Ig be the current flowing through galvanometer.
Let S be the shunt resistance connected in Parallel with galvanometer & Is be the current flowing
through shunt resistance.
From figure,
Ig+Is = I
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Is =I- Ig
P.D. between pts. A & B is same
Ig G = IsS
IgG = (I- Ig)s --------------- 1
S=
Let total current is n times Ig
I= n Ig
S=
S=
(
)
S=
From equation 1
-G = Is - Ig.S
-G + Ig.S = IS
Ig(G+S) = IS
Ig =
(
)
∴ I g+ I s = I
Ig = I - Is
G = S
(I- Is)G = Is. S
IG - Is G = Is S
IG = Is S + Is G
IG = Is(S+G)
Is =
I
Q.9 How M.C.G. can be converted into voltmeter .Derive necessary formula.
Ans:- When M.C.G. is connected in parallel to circuit, then it is called as voltmeter.
Let G be the resistance of galvanometer which gives full scale deflection.
Let Ig be the current flowing through galvanometer & V be the potential difference which is to be
measure .
Let R be the resistance connecting in series with Galvanometer.
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From the figure,
V = I gG + I g R
V = Ig(G+R)
= G+R
R=
−
If V = n Vg
R=
–G
( V=IR,Vg =IgG,G=
R = nG-G
R =G(n-1)
Q.10 Describe construction working & uses of Cyclotron.
Ans:- Diagram
Cyclotron is a device used to produce very high......................
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)
Prof.Anarse.D A
Yashasvi Science Academy
Physics-II
CH.15. Magnetism
Q.1 S.I. current loop produces a magnetic field & behaves like a magnetic dipole.
Ans: -We know, magnetic induction at a point on axis at a distance x from center of circular coil of radius a
carrying current I is given by
B=
(
)
x>> a
∴ Higher order term of a is neglected
B=
(
B=
A=
=
)
B=
Magnetic dipole moment,
M =IA
B=
----------- 1
We know electric field of dipole is given by,
E=
------------- 2
From equation 1 & 2
Magnetic dipole moment M is similar to electric dipole movement P & magnetic field is Analogous to
electrostatic field. Thus, current loop produces a magnetic field & behaves like a magnetic dipole.
Q.2 Derive an expression for magnetic dipole movement of revolving electron
Ans:- diagram
Electron of charge –e perform U cm around a stationary nucleus with period of revolution T. Let r
be the radius of orbit of revolution & V is the orbital velocity of electron
∴ Since, period =
T =
∴ Rate of flow of charge is called as current
I =
I =
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I =
I =
∴ Magnetic dipole moment of revolving electron is given by
M0 = IA
M0=
× r2
M0 =
Q.3 Explain the term gyromagnetic Ratio.
Ans:- The ratio of magnetic dipole movement with angle moment of revolving electron is called as
Gyromagnetic ratio.
.
G.R.
G.R. =
We know, magnetic dipole moment of revolving electron is given by, M0 =
Mult & div RHS by mass of electron (Me)
M0 =
× (MeVr)
L0 = MeVr - angular momentum of revolving electron
M0 =
× (L0)
=
G.R. =
. ×
G.R. = × . ×
G.R. =8.8× 10
c/kg
Q.4 Explain the term magnetism.
Ans:- The net magnetic dipole moment per unit volume is called as Magnetism of..
It is denoted by Mz
Magnetization =
Mz⃗= .
S.I, unit of magnetization
= ]
Dimension of magnetization [
Magnetization is a vector quantity.
Q.5 what is Curies Law.
Ans: - According to curie, magnetization of a paramagnetic sample is directly proportional to external
magnetic field & inversely proportional to the absolute temperature.
Mz ∝
Mz ∝
Combining equation 1 & 2
Mz ∝
Mz =C.
This equation is known as curies law, & c is called as Curie constant.
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Q.6 Explain the term magnetic Intensity.
Ans:- Magnetic intensity is a quantity used in describing magnetic phenomenon in terms of their magnetic
fields the strength of magnetic field at a pt. can be given in terms of vector quantity called as Magnetic
intensity.
B0 = Where, = Permittivity of vaccum
H =Magnetic Intensity
H= nI
The dimension of magnetic intensity is same as dimensions of Magnetism.
Q.7 Discuss magnetization of a ferromagnetic material with the help of Rowland ring.
Ans: - diagram
Magnetization of ferromagnetic material can be studied with the help of Rowland ring (toid Ring)
The material is form into a thin toroidal core of circular cross section. A toroidal coil having ‘n’
turns per unit length is wounded around core & carries current I, the coil is a solenoid bend into a circle.
In absence of ion core magnitude of magnetic field inside the coil is
B0 = Where, = permittivity of vaccum
In absence of iron core magnetic field ⃗ inside the coil is greater than ⃗
B = B0 + BM
Where BM is magnetic field contributed by the iron core.
∴ BM = Mz
B0 = = B = + Mz
B = (H + Mz)
Magnetization & magnetic intensity is mathematically expressed as
Mz = Ψ H
Where, Ψ =
B = (H + Ψ H)
B = (1+ Ψ )
(1+ Ψ ) = =Relative magnetic permeability.
B= H
B= H
Where, = = ((1+ Ψ )
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Q.8 State the properties of diamagnetic substances.
i.
Substances which are weakly repelled by magnet are called as diamagnetic substances.
ii.
When thin rod of diamagnetic substance is freely suspended in uniform magnetic field, then it sets
perpendicular to direction of field.
iii.
When diamagnetic substance is placed in non-uniform magnetic field then it moves from stronger
to weaker part.
iv.
When diamagnetic liquid is placed in a glass & magnetic field is applied then liquid shows
depression in middle.
v.
When diamagnetic liquid is taken in a ‘U’ tube & magnetic field is applied to one arm, then liquid
level in that arm is lowered.
vi.
Magnetic susceptibility is small & negative.
vii.
Diamagnetic substances lose their magnetism on removal of external magnetic field. E.g. Bi, Cu,
Au, Hg
Q.9 State the characteristics of paramagnetic substances.
i.
Substances which are weakly attracted by a magnet are called as paramagnetic substances.
ii.
When thin rod of paramagnetic substance id freely suspend in uniform magnetic field, then it sets
parallel to direction of field.
iii.
When paramagnetic substance is placed in non-uniform magnetic field then it moves from weaker
to stronger part of field.
iv.
When paramagnetic liquid is placed in glass & magnetic field is applied then liquid shows elevation
in middle.
v.
When paramagnetic liquid is taken in ‘U’ tube a magnetic field is applied to one arm then liquid
level in that arm rises.
vi.
The magnetic susceptibility is small & positive.
vii.
Paramagnetic substances lose their magnetism on removal of external field. Hence, they cannot be
used to make permanent magnet.
e.g. Al, Mn, Cr
Q.10 State the characteristic of ferromagnetic substances.
i.
Substances which are strongly attracted by a magnet are called as ferromagnetic substances.
ii.
These substances can be magnetise by external magnetic field only
iii.
Even if external mag field is removing they show magnetism hence, they can be used as permanent
magnet.
iv.
The magnetic susceptibility is positive & high
v.
All properties of paramagnetic substances are true for ferromagnetic substances.
Q.11 Explain Ferromagnetism on the basis of domain theory.
Ans :- diagram
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Physics-II
According to domain theory, ferromagnetic substance contain large number of small regions called as
Domain magnetic Dipole moment of all atoms in 1doamain are ∥ to direction of field. Domain size is 1
mm & it contains 10’’ atoms.
In absence of magnetic field different domains are oriented at random so that magnetic field of
domain cancels with each other. & material does not show magnetic effect.
In presence of weak magnetic field same domains are ∥
to magnetic field .As magnetic field
increases size of such domains increases.so that there is a shift of boundaries between domains, but after
removal of magnetic field boundaries comes to the original position Hence substances loses magnetism.
When strong magnetic field is applied axis of all domain turns in the direction of mag field hence
even if magnetic Field is remove domain axis do not return to its original position. Hence it acts as
permanent magnet.
Q.12 what is Curie temperature?
Ans:- The temperature at which domain structure is destroyed is called as curie temperature.
OR
The temperature at which ferromagnetic substances is converted into paramagnetic
substances is called Curie temperature.
Q.13 Why does a ferromagnetic substance becomes paramagnetic above curie temperature.
Ans:- As temp increases thermal motion of atomic magnet increases & coupling between atomic different
atomic magnets becomes lose. At a certain temperature coupling between every atomic magnet breaks,
hence axis of atomic magnet gets oriented in all possible directions .Hence, ferromagnetic substances turns
to paramagnetic substances.
Problems
Important Formulae
1) I=
2) M=nIA
3)
=
4) B = H
5)
=
6) Ψ ∝
Numerical
Q.1 a circular coil of 300 turns & diameter 14cm carries a current of 15 A. what is the magnitude of
magnetic moment associated with the coil.
An:-n=300
d=14cm
I 15 A
M =?
1) r=
=
= 7cm = 7×10-2 m
2) M=nIA
M = nI( r2)
=300×15×3.142×(7×10-2)2
=3×102×15×3.142× 49 ×10-4
= 45 × 3.142× 49× 10-2
= 2205 × 3.142 × 10-2
= 69.28Am2
Q.2 An electron in an atom revolves around the nucleus in an orbit of radius 0.5 A0.Calculate the
equivalent magnetic moment if the frequency of revolution of electron 1010 MHz.
Ans: - r= 0.5A0= 0.5 × 10-10m
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YASHASVI SCIENCE ACADEMY
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n=1010MH
Yashasvi Science Academy
1010×
106H
1016H
z =
z=
M =?
1) M = IA
= (ne) × r2
= 1016× 1.6 × 10-19×3.142×(0.5× 10-10)2
=1.6×3.142 × 10-3×0.25×10-20
= 1.6×3.142 ×0.25×10-23
=1.257 ×10-23A.m2
Physics-II
z
Q.3 find percentage increase in magnetic field B when a space within current carrying toroid fill
with the aluminium susceptibility of aluminium is 2.1× 10-5.
Ans: - X = 2.1× 10-5
The mag field inside the totoid in absence of magnetic Field, B0 = 0H
When filled with alumninum
B = 0(1+X)H
Increase in magnetic Field
B-B0 = 0(1+X) H - 0H
= 0H + 0 H - 0H
B-B0 = 0 H
% increase in magnetic Field =
=
× 100
× 100
= × 100
= 2.1 × 10-5× 100
= 2.1 × 10-3
Q.4 The space within a current carrying toroid is filled with tungsten with susceptibility 6.8 × 105
.What is the percentage increase in the magnetic field.
Ans:- Χ = 6.8× 10-5
The magneticfield inside the toroid in absence of magneticfield is B0 = H
When filled with tungsten
B = 0(1+X) H
Increase in mag field B-B0 = 0(1+X) H - 0H
= 0H(1+ − 1)
B-B0 = 0H
% increase in magneticfield =
× 100
=
× 100
=6.8 × 10-5× 100
= 6.8 × 10-3
Q.5 A bar magnet made of steel has magnetic moment for 2.5Am2 and a mass of 6.6 × 10-3Kg.If the
density of steel is 7.9× 103 Kg/m3.Find the intensity of magnetization of the magnet.
Ans:- Given M =2.5 Am2
m=6.6× 10-3 Kg
= 7.9× 103 Kg/m3
Mz = ?
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Physics-II
=
I)
=
. ×
V=
V= = density
. ×
.
.
II)
× 10
Mz =
.
=
=
×
. ×
× 10
= 2.999 × 10
= 3 × 10 A/m
Q.6 The susceptibility of magnesium at 300K is 1.2 × increase to 1.8 × Ans:- = 300K
= 1.2 × 10
=?
= 1.8 × 10
∝
At what temp will the susceptibility
=
=
=
×
. ×
. ×
×
=
= 200K
Q.7 the susceptibility of anid iron at saturation 5500 find the permeability of anid iron at
saturation.
Ans := 5500
=?
= 0 r
= 0(1 + )
= 4 × 10 (1 + )
= 4 × 3.142 × 5501 × 10
=12.568 × 5501 × 10
= 6.9 × 10
Q.8. the mag field B & the magnetic Intensity H in a material are found to be 1.6 & 1000 A/m
resp. Calculate the relative permeability & the susceptibility of the material.
Ans:- B = 1.6
H = 1000 A/m
1)
2)
B= H
B=
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H
=
YASHASVI SCIENCE ACADEMY
Prof.Anarse.D A
=
=
III)
×
×
.
.
×
. ×
=
=
Yashasvi Science Academy
.
.
.
= 1.274× 10
= 1+
= -1
= 1.274 × 10 - 1
= 1274 – 1
=1273
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YASHASVI SCIENCE ACADEMY
Physics-II
Prof.Anarse.D A
Yashasvi Science Academy
Physics-II
16 .Electromagnetic Induction
Q.1 Explain the term of Electromagnetic induction.
Ans:- When there is change in magnetic flux, E.M.F. is generated in the coil, so that electric current flows
through coil, This E.M.F. is called das induced E.M.F. & current is called as Induced current.
Production of induced E.M.F. in a coil by changing the magnetic flux linked with it is called as
Electromagnetic induction.
Q.2 State Faraday’s law of electromagnetic induction.
Ans:- I) Faraday’s first law - Whenever there is a change in magnetic flux associated with a coil e.m.f. is
induced in a coil.
II) Faraday’s second law : - Magnitude of induced e.m.f. is directly proportional to the rate of change of
magnetic flux.
Mathematically,
e ∝
e=K .
Where, K= proportionality constant If e=1,
=1
1=K(1)
K =1
∴ e=
Q.3 State Lenz’s law.
Ans:- The direction of induced e.m.f. is such as to oppose the change in magnetic flux which produces it
combining Faradays & Lenz’s law.
e=
Negative sign shows that induced e.m.f. oppose the change in magnetic flux.
Q4. State Fleming’s right hand rule.
Ans:- Stretch the thumb & 1st two fingers of right hand in such way that they are mutually ⊥
to each
other.turn the ahnd such that forefinger points the direction of magnetic field & tumb points the direction
of motion of conductor then the middle finger denotes the direction of induced e.m.f. of conductor.
Q5. What is Eddy’s current (Fo caukt’s current) state & explain application.
Ans:- The circulating current induced in a metal box when it is moving in magnetic fields or place in
changing magnetic field is called as Eddy’s current.
1) Dead beat Galvanometer :- the coil of M.C.G. is wounded on a frame of conducted material like
copper or aluminium when current flows through circuit coil gets deflected but when current
stops coil goes on oscillating but because of Eddy’s current produced in coil, motion of coil can be
prevented & oscillation stops. This type of galvanometer is called as Dead Beat Galvanometer.
2) Induction Furnace : - Heating effect of Eddy’s current is used to melt the metal in the induction
furnace. Metallic solid is placed in a changing magnetic field so that Eddy’s current are developed
in a metal Since, the Eddy’s current are produced on large scale so that heat produced will be large
& metal gets melted. Induction furnace is used for producing alloys of different metals.
3) Electric Breaks: - To stop a train driver cuts off the supply of current to electric motor. But at the
same time, magnetic field is applied to a rotating frame fixed to the axial. The Eddy’s current which
produced oppose the motion & trained stop suddenly.
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4) Speedometer .
Q6. Prove the relation e =
where symbols have their usual meanings.
Ans:- diagram
Consider, rectangle frame PQRS which is kept in uniform magnetic field of induction ⃗ .Magnetic
field is perpendicular to the plane of paper free electrons in the conductor travels in a anticlockwise
direction so that conventional current flows in clockwise direction.
Let I be the conventional current. Consider, a part QR of wire, Let l be length of QR, so that
force F = Bil is acting on part QR according to Fleming’s left hand rule .Direction of force is to the right
of frame .If frame is shifted to the left external mechanical force must be exerted on frame PQRS towards
left.
Mechanical force,
-F = - Bil
Work done by mechanical force is given by,
∴ Work = Force × displacement
dω = -f× dx
dω =- Bil × dx
Let θA = Area swept by rectangular
From Q’QRR’ = l.dx
dω = -BidA
dω = -id ----------------- 1
This work done is converted to energy to maintain current I for time dt if e is induced e.m.f. then work
done also equal to eidt
dω = eidt ---------------------- 2
from equation 1 & 2
eidt = -id e=
e= -BlV
V=
dx = V.dt
We know, Area swept by conductor
dA=ldx
dA = lv.dt
We know change in magnetic Flux
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D = B.dA
=BlVdt
We know ,
e=
e=
(BlV dt)
e= -BlV
Q7. Explain the terms self-conductance & mutual inductance.
Ans:- Self Conductance :- current flowing through coil produces magnetic Field around it when current
changes magnetic field also changes and because of changes in magnetic flux in coil e.m.f. is generated in
coil ,such inductance is called as self-inductance.
Production of induced e.m.f. in coil due to changes of current in the smae coil is called as selfinductance.
Consider, a coil carrying current I.
Let be the magnetic flux of coil
∝ I
= LI
Where L is called as self-inductance which depends upon number of turns of coil.
We know,
e=
=
(LI)
e= - L
If magnitude is considered
∴ e = L
If
∴ e=L
=1
L=
Unit of L =
=
Unit of L = Henry (H)
 One Henry: - self-inductance of the coil is one henry if e.m.f. of 1 volt is induced in a coil when a
current in coil is changes at the rate of 1 ampere per second.
1H=
Dimensions of L =
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=
=[
Physics-II
×
]
 Mutual inductance: Diagram
Two coils A and B are placed closed to each other. Let I be the current flowing through coil A. Therefore it
produces magnetic field around a coil if coil B is kept in this magnetic field.
Magnetic flux is linked with coil if there is change in current in coil A. e.m.f. is induced in coil B this
is called as Mutual Inductance.
Production of induced E.M.F. in one coil due to changes of current in neighboring coil is called as
Mutual inductance. It is denoted by ‘M’.
∴ M=
Q8. With the help of net diagram describe construction & explain working of transformer.
Ans:- Principle:- Transformer is a device with the help of which given alternating voltage can be increase
or decrease to any desired value.
 Construction : Transformer consists of a close laminated soft iron core on which 2 coils having different number of turns
are wound. The coils are made up of insulated copper wire. The coil to which AC input voltage is applied is
called as primary coil & coil across which AC output voltage is obtain is called as Secondary coil.
 Working :When an AC voltage is applied across primary the current through primary changes so that magnetic flux
passing through core changes as this changing flux is linked with both the coils, E.M.F. is induced in each
coil.
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Let N1 & N2 be the number of turns of the primary & secondary coils resp. Let E1 & E2 be the respective
E.M.F.’s induce across 2 coils. Let be the magnetic flux linked with each turn of both coils.
Total flux linked to the primary = N1
Total flux linked to the primary = N2
E1 =
(N1 ) = - N1
E2 =
(N2 ) = - N2
= =
is called Turns ratio.
For ideal transformer, O/P power = I/P power
E2I2 = E1 I1
= =
Q9. What do you meant by Step up & Step down transformer?
Ans: - 1) Step up Transformer: - IT output voltage is larger than input voltage, then transformer is called
as step up transformer
For step up transformer, N > N
∴ E >E
2) Step down transformer: - If output voltage is smaller than input voltage then transformer is called as
Step down Transformer.
For step down transformer, N < N
∴ E <E
Q10. Obtain an expression for E.M.F. induces in a coil rotating in a uniform magnetic field.
Ans:- Diagram
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Suppose a coil is rotated with uniform angular velocity w, about an axis ⊥er to uniform mag
induction B.Let n is the number of turns of coil. A be the area of coil.suppose in time t coil rotates from
position PQ to position P’Q’ making an angle θ between them.
∴ θ = ωt.
After time t, when coil is at position Q’P’ mag flux passing through coil is,
= nAB cosθ
We know,
e=
e=
e=
(nAB cosθ)
(nAB cos ωt)
e= n ABω sin wt
But w =2 f
e = 2 fnAB sin wt
Let = 2 fn AB = Peak (greatest) value of E.M.F.
e= sin ωt
Diagram
Q11. Explain Simple AC current with resistance only.
Ans:-
Consider, an alternating source of E.M.F. applied between terminals of resistance R.
The value of alternating e.m.f. at nay instant is given by
e= sin ωt
The current I in resistance R is given by,
e=IR
I = I=
Let
=
I=
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= Peak value of current
sin ωt
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Therefore, current is in phase with E.M.F..
Q12. Explain what do you mean by RMS value of alternating e.m.f. & current. How are they related
to their peak values?
Ans:- Let be the peak value of alternating current flowing through resistance R. Let current is flowing
for time t & heat produce is H. If same quantity of heat is produced then same resistance R in same time t
by passing steady current of constant magnitude then this steady current is called as RMS value/effective
value /Virtual value.
This current is denoted by
or I
It is given by
=I=
√
Similarly RMS value for e.m.f. is,
=E=
√
Q13. Derive the relation for inductive reactance.
Ans:- Diagram
Let alternating E.M.F. is applied across inductor of inductance L Let I be the instantaneous current
flowing through inductor.
I=
sin ωt ----------- 1
E.M.F. is induced in inductor & it is equal to instantaneous value of applied E.M.F.
e= L
e= L. [ sin ω ]
e= L ω cos ωt
e= L ω sin (ωt + )
Let
= L wPeak value of e.m.f.
e= sin(ωt + ) ----- 2
From equation 1& 2
Current logs behind the e.m.f. by
=L ω
= Lω
√
√
= ωL
= ωL
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This equation is similar to Ohm’s law; RMS is resistance of inductor which is called as Inductive Reactance.
It is denoted by XL
∴ XL = ωL
XL = 2 fL
Q14. Derive an expression for capacitive reactance.
Ans:- Let alternating e.m.f. is applied across a capacitor of capacitance C. The instantaneous current across
the capacitor is given by
i= sin ωt ------------------------- 1
Let q be the charge on capacitor. Therefore voltage drawn across capacitor is given by
e= ----------------- 2
i=
dq= i.dt
Integrating
∫ dq = ∫ i.dt
q=∫ sin ωt.dt
q= q=
cos ωt
sin (ωt - )
sub. This value in equation 2
e=
Let
=
sin (ωt - )
= Peak value of e.m.f.
e=
sin (ωt - ) ---------- 3
From equation 1 & 3current leads the e.m.f. By
∴
=
=
√
=
√
=
This equation is similar to Ohm’s law .RMS is resistance of capacitance .Hence called as Capacitive
reactance .It is denoted by Xc
Xc =
Xc =

Capacitive reactance (definition) : - It is the ratio of RMS voltage across capacitor to the RMS
current passing through it.
 L-C-R series circuit:Diagram
e=
sinωt
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e=
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I = sinωt
sin(ωt+ )
+(
z=
)
−
= impedance of
Cos =
(
)
= power factor =
 For resonance circuit:1) Conditions for series resonance is,(acceptor C++)
=
ωL =
ω =
ω=
√
fr =
√
fr =
√
2) For parallel resonance min current flows but formula remains same
fr =
√
Power in ckt with resistance only,
P=V × I
P=e ×i
P= sinωt ×
sinωt
Integrating this relation for one cycle.
P=∫
ωt
P=
∫
ωt
∫
wt over a complete cycle =
P=
P=
√
×
×
×
√
P=
×
 Power in L-C-R series circuit
e=
sinwt
i= sin(ωt± )
P=e× i
P=
sinwt × sin(ωt± )
P=
sin wt. sin(ωt± )
P=
ωt cos
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wt over a complete cycle =
∴P=
P=
P=
Where cos
cos
z=
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× × cos
√
×
√
× cos
×
× cos
is called as Power factor.
=
+(
−
)
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17. Electron & Photon
Q.1 What is photoelectric effect?
Ans:- The phenomenon of emission of electrons by metal surface under the action of light is called as
photoelectric effect.
Q.2 With the help of neat circuit diagram describes exp. to study photoelectric effect.
Ans:- Diagram
Construction :- The apparatus consist of evacuated glass tube ‘G’ having quartz window ‘W’ Two metal
plates emitter E and collector C are fixed in tube. They act as two electrodes, emitter plate is made up of
material whose photoelectric emission is to be studied.
UV light from source of variable frequency & intensity enter tube through window W. collector collects
photoelectrons battery supplies P.D. upto 10v. to collector P.D. can be change by potential divider
arrangement. Plate C can be give positive potential by means of commutator.
Characteristics:1.
Initially plate C has given positive potential & UV light of known intensity and low frequency is
allowed to fall on plate E. Increase the frequency by keeping intensity constant at low frequency,
current does not flow but as soon as the frequency reaches certain value current starts flowing.
This equation is called as Threshold Frequency. It is denoted by ( )
2. Keep frequency of source greater than threshold frequency & increase the intensity of light .As
intensity increases photoelectric current also increases .i.e. rate of emission of photoelectrons is
directly proportional to intensity of incident radiation.
3. Keep frequency & intensity of source constant & decrease positive potential applied to collector.
Photoelectric current go on decreasing but even if potential difference applied is zero. small current
flows through circuit. then reverse commutator key so that negative potential is applied to
collector plate. negative potential oppose the electrons to flow towards collector so that current
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flow stops. Magnitude of retarding potential at which photoelectric current is 0. is called as
stopping potential.
4. Make intensity of source very low photoelectrons are emitted continuously .This shows that
photoelectric emission is an instantaneous process
Q.3 State the characteristics of photoelectric effect.
i.
ii.
iii.
iv.
Photoelectric are emitted only if frequency of certain minimum frequency called as Threshold
frequency. The wavelength corresponding to Threshold frequency is called threshold.
Rate of photoelectric emission is directly proportional to intensity of incident radiation.
The maximum kinetic energy with which photoelectrons are emitted depends upon frequency of
incident radiation but it is independent of intensity of radiation.
Photoelectric effect is an instantaneous process i.e. there is no time lag incidence of radiation &
emission of photoelectrons.
Q4. Describe how Einstein proved different characteristics of photoelectric effect.
Ans:- Einstein’s photoelectric equation is –
ℎ =
+ω
m
Where, h = Plank’s constant
V = Frequency of radiation
=
m
= max K.E.
= Photoelectric work function

Photoelectric Work function : -Amount of energy required to removed the electrons from the
surface of metal is called as photoelectric work function.
i) When frequency of incident radiation decreases kinetic energy of photoelectrons also decreases .Let it
becomes 0 when frequency is V0
When V = V0
m
=0
∴ h V0 = 0+ ω
∴ ω = h V0
hυ =
m
+ h V0
hυ - h υ0 =
h (υ- υ0) =
m
m
From this equation ,if υ > υ0 photoelectrons are emitted & if υ < υ0, photoelectrons are not
emitted.
ii) If intensity of beam increases it contain more no of photons. Therefore, there are more collisions
between photons & electrons .Hence, rate of photoelectric emission increases.
iii) Einstein’s photoelectric equation is
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hυ = m
+
ω is constant for same emitor hence if frequency increases max K.E.
increases & it is independent of intensity of radiation.
iv) Photoelectrons are emitted if there is collision between photons & electrons. As soon as radiation is
incident collision takes place & photoelectrons are emitted, hence it is an instantaneous process.
Q5. What is photoelectric cell. Describe its construction and working.
Ans.:- A cell which converts light energy into electrical energy is called as
photoelectric cell.


Construction:- Photoelectric scale consist of evacuated tube in which the electrodes cathode(K) &
Anodes(A) are fixed. Cathode is cylindrical in shape .Concave surface of cathode is coated with the
photosensitive material like sodium. Anode is a platinum rod. glass tube is fitted on a non
conducting base. Cathode and anode are internally connected by pins. Cathode is connected to
negative terminal & anode is connected to positive terminal. Milliammeter is connected in series
with anode.
Working:- When intensity radiation increases, the rate of photoelectric emission increases &
current is flowing through circuit ,thus, current is directly proportional to intensity of incident
radiation.
Q6. State the application of photoelectric cell. Explain any one of its application.
Ans.:- Applications:-1.Exposure meter 2.Lux meter 3.Burglar alarm In production of sound.
i)
ii)
iii)
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Exposure Meter:- It is used in camera to find correct time for having good exposure.
Photoelectric cell is enclosed in a case having 2 openings through one opening light is allow to
fall on photoelectric cell & through other opening milliammeter can be observed when light is
poor deflection is small when light is bright deflection is large. In poor light camera has to keep
for longer time to obtain a good photograph.
Lux meter :- It is used to measure intensity of light .It is same as the exposure meter but scale
is calibrated in terms of lux.
Burglar Alarm:- It is used to save the precious things from Burglar. Photoelectric cell is
connected in close circuit infrared source of light is incident so that current is continuously
flowing through circuit. When Burglar approach towards cell. Infrared radiations are
intercepted & photoelectric current stops. Relay is operated & electric current send to another
circuit containing electric bell thus, bell starts the ringing & alarm raised
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18 Atoms, Molecules & Nuclei
Q.1 State & Explain Bohr’s law.
i.
The electrons in Hydrogen atom revolves in a circular orbit around the nucleus situated at the
centre of orbit.
ii.
Centripetal force required for this circular motion is provided by the electrostatic force of attraction
between positively charge nucleus & negatively charge electrons positivebe the charge on nucleus &
negativebe the charge on electron. Let r be the radius of orbit then by Coulomb’s law Electrostatic
force of attraction exerted by nucleus on electron is given by ,
Electrostatic Force = Let m be the mass of electron & v be the velocity of electron then centripetal force acting on
the electron is given by ]
Centripetal force = According to Bohr’s 1st Postulate : =
Q.2 State & Explain Bohr’s 2nd Postulate.
Ans:-Statement :- Electron can revolve without radiating energy ,only those orbits for which the angular
momentum of electron is equal to an integral multiple of
L = n
Where, h = Planck’s constant, n = Principle quantum no.
Moment of inertia of electron about its axis of rotation is given by
I = mr2
v = rω
ω=
L = Iω
L = mr2 ×
L= mvr
nd
According to Bohr’s 2 postulate,
Mvr = n
Q.3 State & Explain Bohr’s 3rd postulate.
Ans:- An electron radites energy only when it jumps from higher energy orbit to lower energy orbit. The
energy radiated is equal to difference between energies of the electron in the 2 orbits. This energy is
radiated in the form of photon of energy h
where h = Planck’s constant
= frequency of radiation
Let En be the energy of electron in the higher energy orbit & Ep be the energy of electron in lower energy
orbit. then according to Bohr’s 3rd postulate.
En - Ep = h
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Q.4 S.T. radius of Bohr’s orbit is directly proportional to square of
principle quantum number.
Ans:- According to Bohr’s 1st postulate
C.F =E.F.
=
=
×
=
---------------- 1
According to Bohr’s 2nd postulate
mvr= n
v= ------------------------- 2
=
---------------- 3
From 1 & 3 ,
=
=
×
r=
∴
is constant
∴r∝
Thus, radius of Bohr’s orbit is directly proportional to square of principle quantum no.:From equation 2, v=
v=
∴v∝
v=
Q.5 Prove that energy of Bohr’s orbit is inversely praportioanl to square of
principle quantum number.
Ans:- According to Bohr’s 1st law postulate
C.F.= E.F.
∴
=
=
×r
=
K.E. =
Potential at distance r from charge e
V=
This potential represents P.E. of unit mass charge
∴ P.E. of electron of charge(-e)
P.E. = v(-e)
P.E. =
× (-e)
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P.E. =
Physics-II
Total energy is the sum of K.E. & P.E.
E. = K.E.+ P.E.
=
-
=
=
=
.
E =
Is constant
∴ E∝
Thus, energy of Bohr’s orbit is inversely Proportional Square of principle quantum no.
.
Remark: =
eV
= -13.6 eV
.
=
= 3.4ev
.
=
= -1.51ev
.
=
= -0.85 e
Q.6.Explain the origin of spectral lines hence draw the energy level diagram of hydrogen atom.
Ans:- Let En be the energy of electron in the nth orbit. & Ep be the energy of electron in pth orbit.
Therefore,
En =
Ep =
.
∈ .
∈ Let n > p
En - Ep =
=
.
-
∈ .
.
∈ −
∈ According to Bohr’s 3rd Postulate,
En - Ep = h
H =
=
=
=
.
−
∈ .
−
∈ .
−
∈ .
−
∈ Let R =
.
∈ =Rydberg’s constant.
= 1.097 × 10 /m
= R −
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i) Transition of electron from outer orbit of p=1 to n=2,3,4,...................... gives Lyman series,
= R −
, n= 2,3,4.....
All the wavelength of Lyman series lie in the ultraviolet region.
ii) Transition ................ p=2 to n =3,4,5,..............
gives Balmer series for Balmer series
= R −
n = 3,4,5...
If p = 2 ,n=3 then it gives
Line
If p = 2 ,n=3 then it gives
Line
If p = 2,n=5 then it gives
line.
All the wavelength lie in visible region.
iii) Transition ................. p=3,n=4,5,6,........... gives
Pashcen series For paschen series.
= R −
n = 3,4,5...
All the wavelength lie int the infrared region.
iv) Transition ............. p=4,n=5,6,7,......... gives Brackett series .For Brackett series
= R −
n =5,6,7,.....
Wavelength lie near infrared region.
v) Transition ............... p=5,n=6,7,8,.............. gives p funds series .For Pfnds for Pfunds series
=R
−
n = 6,7.,8........
Wavelength lie in the infrared region.
Q.7 State & Explain de Broglie’s Hypothesis.
Ans:- According to de Broglies hypothesis ,a matter particle of momentum p has a wave associated if
with.....having wavelength given by = where h is Planck’s constant.
Let m be the mass of particle & v be the velocity of particle.
=
According to Planck’s theory E = h ---------- 1
By Einstein’s mass-energy relation E = mc2 ---------2
From equation 1 & 2
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∴h =
= mc2
mc2
=
=
=
E=
= 2Em
p= √2 =
√
Q.8 Calculate wavelength of electron by de Broglie’s hypothesis.
Ans:- Consider an electron of mass me accelerated through a P.D. V volts. Let
Be the velocity of electron & e be the charge on electron .The energy gain by electron moving through
P.D. of volts is eV.
∴ Electrical energy =eV
K.E. =
=
∴ KE = electrical Energy
= eV
= 2 eV
P= 2 eV
By de Broglie’s hypothesis,
=
=
.
=
.
×
√
=
A0
√
Q.9 Describe Davisson and Germer Experiment.
Ans:- diagram
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i) Davisson & Germer perform the expt to study wave nature of electron

Construction:- Expt. consist of Tungsten filament when current is pass through filament. It
becomes hot & emit electron hence filament act as Cathode .These electrons are accelerated towards
anode because it is at positive potential. The P.D. between Anode & Cathode can be change by
potential divider arrangement. After emerging from hole of anode electron fall on Nickel crystal
there is a detector to receive reflected electron beam & to measure its intensity. The detector can be
rotated to diff. positions.
 Working:- When electron beam is incident normally on nickel crystal then it produces the
diffraction from layers crystal the detector is rotated in different positions & intensity of reflected
beam is measured.
The electron beam is scattered from different atomic planes of crystal .It is found that
intensity of reflected beam of electron is maximum at angle of = 500 and accelerating of
54V.
∴ θ + + θ = 1800
2θ+
= 180
2 θ + 50 = 180
2 θ = 130
Θ = 650
According to Bragg’s law, = 2dsin θ
Where d= distance between 2 atomic planes
=0.91 A.U.
= 2 × 0.91 × 10-10 sin 65
= 1.65 A0 ------------- 1
By de Broglie’s wavelength,
.
=
=
.
√
√
= 1.67 A0 ------------------ 2
From equation 1 & 2 experiment & theoretical values of wavelength of electron are nearly same which
confirms wave nature of electrons.
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