WAVE OPTICS
(IMPORTANT CONCEPTS)
Branch of physics based on the wave concept of light is called ‘wave optics’ or ‘physical optics’.
Wave front:-A wave front is defined as the continuous locus of all the particles of a medium, which are
vibrating in the same phase.
Depending upon the shape of source of light wave front can be of different types.
i)Spherical wave front:-source of light is a point source
eg:-
ii)Cylindrical wave front:- source of light is linear.
iii) Plane wave front:-When the point source or linear source is at very large distance, a small portion of
wave front appears to be plane which is called as planewavefront.
HUYGEN’S PRINCIPLE:-
1) Every point on the given wave front acts as a fresh source of new disturbance, called secondary
wavelets which travel in all directions with velocity of light.
2) A surface touching these secondary wavelets tangentially in the forward direction at any instant gives
the new wave front at that instant. It is called secondary wave front.
a) LAW OF REFLECTION (HUYGEN’S PRINCIPLE)
From the figure
In ∆ADC &∆ AEC
∟D=∟E= 90˚
;AC=AC= Common side
AE=DC=CT
By R.H.S Congruency these two Δ΄are congruent
So CPCT are equal
∟DAC=∟ACE
I=r
Angle of incidence = angle of refraction
b) Law of refraction: Huygen’s Principle
From the figure we have
AD=V2 t ; BC=V 1t
According to definition of law of refraction,
n2 / n1 =sin i/ sin r
From the ∆ABC Sin i= BC/AC
In the ∆ADC Sin r=AD/AC
n2 / n1 =sin i/ sin r = BC/AC * AC/AD= V 1t/ V2 t= V 1/ V2
n2 / n1 =sin i/ sin r= V 1/ V2
Huygen’s principle obey the law of refraction
RELATION BETWEEN PATH DIFFERENCE &PHASE DIFFERENCE
λ IS The path difference
Then phase difference is 2Π
If x is path difference
i.e. x is λ
φ is 2Π
φ = 2Πx/ λ
x= φλ/2Π
COHERENT SOURCES:
The sources of light which emit continuous light waves of the same wave length, same frequency
and in same phase or having a constant phase difference are called coherent sources
Two independent sources of light cannot be coherent
INTERFERENCE:
Two waves super imposed each other as a result the intensity of resulting wave increase or decrease
is called “interference”
Interference are two types. They are
1) Constructive interference: the crust and crust, trough and trough of two waves are
superimposed; as a result maximum intensity wave is formed
2)
Destructive interference: when two waves are super impose crust of one wave combined with
the trough of another wave and the intensity of resulting wave decreases
When there is a modification in energy distribution because of super imposing of waves is known as
“interference”(I)
I is directly proportional to A2
I=a12+ a22+ 2 a1 a 2 cosφ
Imax=( a1+ a2)2
φ=0
Imin=( a1- a2)2
φ=180°
I=I1+I2+2√ I1 I2 cos φ
Imax/ Imin=( a1+ a2)2 /( a1- a2)2= (r+1) 2/(r-1) 2
Intensity of slit is directly proportional to width of the slit → I ά W
I1/I2= a12/ a22=W1/W2
I=a12+ a22+ 2 a1 a 2 cosφ
a1= a2 = a
I=2a2+2a2 cosφ
I=2a2 (1+ cosφ)
I=2 a2 X 2 cosφ/2
I=4 a2 cosφ/2
CONDITIONS FOR BRIGHT BAND:
It is formed during constructive interference.
Path difference φ = 0°,2Π,4Π,6Π,……2nΠ
Phase difference =δ =0; λ; 2 λ; 3 λ;…; n λ
CONDITION FOR DARK BAND:It is formed due to destructive interference.
Path difference φ = Π,3Π,5Π,…(2n+1)Π
Phase difference: δ = λ/2, 3λ/2… (2n+1) λ/2
YOUNG’S DOUBLE SLIT EXPERIMENT:
In young’s doubles slit experiment 2 narrow slits s1ands2 separated by distance‘d’act as coherent
sources.screen is placed at a distance ‘D’ from slits. coherent waves come out from the two slits. These
are involved interference and alternatively dark and bright band are formed on the screen. At P there is
always bright band which is called as central band.on either sides alternative bands are formed.
1)position of bright fringes is given by x=n λD/d where n=1,2,3,for
2) Position of dark fringes is given by x = (2n+1) λ/2 D/d where
N=1, 2, 3,…for 1st ,2nd ,3rd dark fringes.
3) Width of each bright / dark fringe is β=λD/d
1st ,2nd ,3rd …bright fringes.
4) When the entire apparatus is immersed in a transparent medium of refractive index μ, fringe width
β'=λ'D/d=λD/μd=β/μ
5)angular width of interference fringes Ө=β/D=λ/d
NUMERICALS FOR SUPPORTIVE LEARNERS
1Q)Two slits in young's double slit experiments are illuminated by two different sodium lamps
emitting light of same wave length.do you observe any interference pattern on the screen?
Ans: No, interference pattern is not obtained. This is because phase difference between the light
waves emitted from two lamps will change continuously.
2Q)Why is interference pattern not detected ,when the two coherent sources are far apart?
Ans: As fringe width β ά 1/dTherefore when d is so large the width may reduce beyond the visible
region.hence the pattern will not be seen.
3Q)No interference pattern is detected when two coherent sources are infinitely close to each
other why?
Ans: when d is negligibly small, fringe width β which is proportional to 1/d may become too
large.Even a single fringe
may occupy the screen. hence the pattern cannot be detected.
4Q) Consider interference waves from two sources of interferences I and 4I.Find interference at
points where phase difference is i) Π/2 ii) Π
Ans: As we know
2
2
2
A = a1 + a2 + 2
a1 a 2 cosφ
Ia=I1+I2+2√I1√I2
When φ= Π/2
Ia=I+4I+2√4I2(cos Π/2)
Ia=5I
When φ= Π
Ia=I+4I+2√4I2(cos Π)
Ia=5I-4I
Ia=I
5Q)In YDE the intensity of central maximum is I,what will be the intensity at the same place if the
slit is closed?
Ans: When one slit is closed, amplitude becomes ½ and hence intensity becomes ¼ and there is no
interference.
6Q)Widths of two slits in young’s experiment are in the ratio 4:1. what is the ratio of their
amplitudes of light waves from them?
Ans: W1/W2=I1/I2= a12/ a22=4/1
a1/a2=2/1
a1:a2=2:1
7Q) what are coherent sources of light? why no interference pattern is observed when two
coherent sources are i)too close ii)very far apart
Ans:The sources of light which emit light waves of same wave length, same frequency and in same
phase are having a constant phase difference are called coherent sources .
1) when the sources are too close, distance (d) between them tends to zero. Fringe width β=λD/d
tends to ά. Hence no interference pattern will be observed.
2)when the distance (d) between them becomes large fringe width β becomes too small and no
longer be in the visible region. hence no interference pattern will be observed.
8Q) State the condition which must be satisfied for two light sources to be consent.
1. Coherent sources of light should be obtained from a single source by some device.
2. The two sources of should give mono chromatic light.
3. The path difference between light waves from two sources should be small.
Q9) The two slits in YDE are separated by distance 0.03m. when light of wave length 5000A 0 falls
on the slits an interference pattern is observed on the screen1.5m away. Find the distance of
fourth bright fringe from the central maximum?
Ans: d=0.03m
λ=5000 X 10-10
D=1.5m
N=4
x=n λD/d
= 4 X 5000X10-10X1500/30
=106-10-10
=10-4=0.1mm
10Q)In young’s experiment, the width of the fringes obtain with light of wave length 6000A 0is 2.00
mm what will be the fringe width is the entire apparatus is immersed in a liquid of n=1.33?
Ans: β= λD/d
β'=λ'D/d
β'/β = λ'D/d X d/ λD
= λ'/ λ=1/n
β'= β/n
= 2.0/1.33
= 1.5 mm
11Q)In Y.D.E. the slits are 0.03cm apart and these screen is placed 1.5 m away.the distance b/w
the central fringe and 4th bright fringe is calculate the wave length of light is used?
Ans: d=0.03cm = 3X10-4m
x=1cm=10-2m
n=4
D=1.5 m
λ=?
For bright fringes x=n λD/d
λ=xd/nD = 10-2X3X10-4X102/4X1.5
λ=0.5X10-6m
λ=5X10-7m
12Q) In Y.D.E the width of the fringes obtained with the light of wavelength 6000A0 is 2.0mm what
will be the fringe width of the entire apparatus is immersed in a liquid of n=4/3?
Ans: β = λD/d
When the entire apparatus is immersed in water.
λ'= λ/n
β'= β/n = 2.00/4/3 = 1.5mm
HOT’S
1Q) The width of one of the slits is a Y.D.E is double of the other slit. Assuming that the amplitude
of the light coming from a slit is proportional to the slit width, find the ratio of maximum intensity
in the interference pattern.?
Ans: Let ‘A’ be the amplitude of light from one slit
. . . Amplitude of light from 2ndslit is 2A.
Imax/Imin =(a1+a2)2/(a1-a2)2 = (2A+A)2/(2A-A)2 = 9A2/A2 = 9:1
2Q) A double slit is illumated by light of λ=6000A0 the slits are 0.1cm apart and the screen is placed 1m
away. Calculate.1)angular position of 10th maximum in radian.2)separation of two adjacent minima?
Ans: λ=6000A0 = 6x10-7
d=0.1cm = 10-3 ;D=1m
n=10
x=n λD/d
if Ө is angle of diffraction then sin Ө=x/D=n
λ/d=10X6X10-7/10-3
=6X10-3
As sin Ө is very small Ө≈sin Ө = 6X10-3rad
2) separation of two adjacent minima
β=λD/d = 6x10-7x1/10-3 = 6x10-3
3Q) I n young’s double slit experiment , two coherent sources are 1.5m apart and and the fringes are
obtained at a distance of 2.5m from them. If the sources produce light of wavelength 589.3 nm ,find
the no. of fringes in the interference pattern which is 4.9x10-3 long?
Ans: Here d=1.5 nm= 1.5x10-3 m.
D=2.5m, λ=589.3nm=589.3x10-3
fringe width β= λd/D
589.3x10-9x2.5/1.5x10-3
=982.17x10-6m
No of fringes X/β
=4.9X10-3/982.17X10-6
=4.99 ≈5
4. The interference frings for sodium light (λ=5890 AO) in a double slit experiment have an angular
width oof 0.2AO .For what wave length will the width be 10% greater?
Ans: as angular directly proportional to λ
λ’/λ=ө’/ө
now ө=0.20 = Ө’ = 0.2+10(0.2)/100 = 0.22
λ = 5890A0
λ’ = ө’/ө.λ = 0.22x5890/0.2 = 6479A0
5.Q)Two slits are made 1mm apart and the screen is placed 1m away. What is the fringe separation
when blue green light of wavelength 500nm is used?
Ans: d = 1mm = 10-3m
D = 1m λ= 500nm = 500x10-9m = 5x10-7m
Fringe separation = fringe width
β= λD/d = 5x10-7x1/10-3 = 5x10-4m
β = 0.5mm
6,Q)In aY.D.E λ = 500mm and D = 1.0m the minimum distance from the central maximum for
which the intensity is half of the maximum intensity?
Ans: IA =4 I0cos2 ¢/2 = ½ (4I0)
Cos2¢/2 = ½ cos¢/2 = 1/√2 ¢/2 = 450
¢ = 900
Minmum distance from central maximum corresponding to ¢ = 900 would be ¢/2π x β
X = π/2/2π x λD/d
X = ¼ x 500 x 10-9x 1/1 x 10-3 = 1.25x10-4m=
7Q) In YDE while using a source of light of wavelength 5000AO, the fringe width obtained is
0.6 cm.If the distance between the screen and the slit is reduced to half , what should be the
wave length of the source to get fringes 0.003 wide?
Ans:
λ=5000 AO=5000 x 10-10 m
P1 =0.6 cm=0.006 m
D1=D (say)
λ 2=?
β 2=0.003
D2=D/2
Let‘d’ be the distance between b/w the slit is the two cases
β 1= λ1D1/d1
or,
λ2=
β2= β 1D2/d2
β1/ β1 = λ1D1/ λ2D2
λ1D1 β2/ β 1O2
λ2 =5000X10-10XDXDX0.003/0.006XD/2
λ=5X10-7m
8 Q) Find the maximum intensity in case of interference of n identical waves each of intensity I O if the
interference is a)coherent b)incoherent
Intensity of each of the ‘n’ waves = I0
When interference is due to coherent sources
I=I1+I2+2√I1I2cos ¢
Imax= I1+I2+2√I1I2 cos 0°
= (√ I1+√I2)2
For n identical waves each of intensity (I0)
Imax=(√ I0+√ I0+… n times) I0
=(n√ I1 + √ I2)2 = n2 I0
When the interference is due to incoherent source ¢ varies randomly with time.
(cos¢)av= o
Imax= I1+I2
For n identical waves each of intensity IO
Imax= IO+IO+……… ..(n times)
Imax =n IO
9Q) In young’s experiment two slits are 0.2mm apart the interference fringes for light of
wavelength 6000A0 are formed on a screen 80cm away
a) How far is the 2nd bright image from the central image?
b) How far is the 2nd dark image from the central fringe?
Ans: Here d = 0.2mm = 2 x 10-4m
λ = 6000A0 = 6 x 10-7m
D = 80cm = 0.8m
a)
X=?
n=2
x = nλD/d
=
=
2 x 6 x 10-7 x 0.8/2 x 10-4
4.8 x 10-3m
b)
X =?
n=2
x = (2n+1) λD/2d
= 5 x 6 x 10-7 x 0.8/2 x 2 x 10-4
= 6 x 10-3m.
10Q) Green light of wavelength 5100A0 from a narrow slit is incident on a double slit. If the
overall separation of 10 fringes on a screen 200cm away is 2cm, find the slit separation?
Ans:
λ = 5100A0 = 5.1 x 10-7m
D = 200cm = 2m,
d =?
As β = λD/d
d = λD/β = 5.1 x 10-7 x 2/2x 10-3
= 5.1 x 10-4m
DIFFRACTION
(Important formulae and Concepts)
Diffraction of light is the phenomenon of bending of light around corners of an obstacle or aperture
in the path of light on account of this bending light penetrates in to the geometrical shadow of the
obstacle.The light does deualts from it’s linear path.
The diffraction pattern due to a single slit consists of a central bright band having alternate dark and
week bright bands of decreasing intensity on both sides .
Condition for diffraction maximum.
a sinө = (2n+1) x λ/2 where n = 1, 2,3,…………………
Condition for diffraction minimum.
a sinө = n λ where n = 1,2,3,4,5,………………………
width of the central maximum
2x = 2x λf/a =
2x λ D/a =2β
Angular width = 2Ө
2 x λ /a= 2 x/f = 2x/D
Diffraction is supposed to be due to interference of secondary wavelets from the exposed portion of
wave front from the slit
Where as interference, all bright fringes have same intensity, in diffraction bright bands are of
decreasing order intensity
Differences between interference & diffraction
Interference
Diffraction
1)Two coherent waves superimposed each 1)Different wavelets of the same wave
other as a result interference pattern is superimposed as a result diffraction is
observed
obtained
2)width of all dark bands are equal as well 2)width of the central band is maximum
as bright bands
3)Intensity of all bright bands are equal
3)intensity of central band is maximum
&decreases with increasing order
4) conditions for interference maximum is 4)conditions for diffraction maximum is
a
a sin Ө=(2n+1)λ/2
¢ =2nπ
δ=nλ
5)diffraction minimum a sinӨ=nλ
5)conditions for interference minimum
¢ = (2n+1)π δ= (2n+1)λ
6) intensity distribution graph
6)intensity distribution graph
Fresnel distance :
It is the minimum distance a beam of light has to travel before it’s deviation from straight line path
becomes significant
Where Zf = a2/ λ
Zf →
fresnel distance
POLARISATION:
Unpolarised light:
When light is come out from the source it will move in all possible direction with equal velocity i.e.
particles vibrates in all directions is known as unpolarised light
If light particles vibrate in unidirectional it is called as polarized light.
→ Unpolarised light can be converted into polarized by using polarisers
When the polarizer is placed parallel in front of the incident unpolarised light it will allows the light
particles which are parallel to the direction of slit of the analysers is placed parallel to the polarisers
same pattern is observed. Intensity of light come out from the analyser equal to intensity of light
incident on analyser
If analyser is placed perpendicular to polariser nolight come out from the analyser
MALUS LAW:
Intensity of light come out from the analyser is directly propotional to the square of cos angle
I is directly propotionat to cos2Ө
I=I0 cos2 Ө
Where Ө is an angle between polariser and analyser
I0→ intensity of light which is transmit through the polariser
Intensity vs Ө graph
POLARISATION OF LIGHT BY REFLECTION:
From the figure we have
Ip+r2p=90o
r2p=90o-ip
A/c to def of refractive index
n = sin i / sin r
Sin ip/sin (90-ip)= n
n=sin ip/cos ip=tan ip
n= tan ip
Refractive index is directly proportional to tan of polarizing angle
USES OF POLAROIDS:
1.To avoid glare of light
2.to avoid dazzling light of a car approaching from the opposite side during night driving
3.In three dimentionals motional pictures in halography.
4. To improve colour contrast in old oil paintings.
5. Used in a optical stress analysis.
NUMERICALS FOR SUPPORTIVE LEARNERS
1. Name one device producing polarized light?
Nicole prism or calcite prism.
2. What is the value of refractive index of a medium of polarizing angle 600?
N=Tanip
Tan 600=3
N=3
3. What does red shift in the spectra indicates?
The universe is expanding.
4. Diffraction is common in sound but not common in light waves. Why ?
For diffraction of a wave, an obstacle or operture of the size of the wave length of wave is
needed. As wave length of light is the order of 10-6 m and obstacle of this size is rare, therefore
diffraction is not common in light waves. On the contrary, wave length of sound is the order of 1
meter and obstacle of the size is readily available. Therefore diffraction is common in sound.
5
In a single slit diffraction the width of the slit of the made double the original
width. How does this effect the size and intensity of these central diffraction
brand?
Width of the central maximum is 2x=2D/a. when width of the slit a is doubled
central maximum width is halfed. Its area becomes ¼ th . hence intensity of
central diffraction brand becomes four times.
6. What evidence is there to show that sound is not electromagnetic in nature?
Sound waves are not electromagnetic in nature. This is evident from the fact that sound waves
are require a material medium for propagation. For example, when a jar fitted with an electric
bell is evacuated no sound is heard though hammer is seen to strike the bell. This is because
without air sound can not travel.
7. Can a naked eye detect polarization of light? If not how is polarization of light detected?
No. a naked eye can not detect polarization of light. This distinction can be made
by using a crystal. A calcite crystal , quartz crystal , a Nicole prism can be used a
polarizer as well as analyzer.
8. Two narrow slits are illuminated by a single monochromatic source. Name the pattern
obtained on the screen. One of the slit is now completely covered , what is the nature of the
pattern obtained now on the screen? Draw intensity pattern obtain in the two cases? Also
write two differences between the patterns obtained in the above two cases.
When two narrow slits are illuminated by a single monochromatic sourece, the
pattern obtained on the screen is interference pattern consisiting of alternative
bright and dark frings. When one of the cities covered completely no interference
occurs. When we obtain is diffraction pattern due to single slit.
Intensity pattern in the two cases are shown.
Diagram.
Interference:
1. Two coherent waves superimposed each other as a result interference pattern is
observed
2.width of all dark bands are equal as well as bright bands
3.Intensity of all bright bands are equal
4. conditions for interference maximum is a
¢ =2nπ
δ=nλ
5. conditions for interference minimum
¢ = (2n+1)π δ= (2n+1)λ
6. intensity distribution graph
Diffraction:
1) Different wavelets of the same wave superimposed as a result diffraction is obtained
2) width of the central band is maximum
3) intensity of central band is maximum &decreases with increasing order
4) conditions for diffraction maximum is a sin Ө=(2n+1)λ/2
5) diffraction minimum a sinӨ=nλ
6) intensity distribution graph
9. The light of wave length 600 nm is incident on an aperture of size 2mm. calculate the distance
upto which the ray of light can travel, such that its spread is less than size of aperture.
Here =600 nm = 600X10-9 m
A=2 mm = 2X10-3m, ZF = ?
We know ZF = a2/ = (2X10-3)2/600X10-9 = 6.67 m.
10. Light of wave length 600 nm is incident normally on slit of width 3 mm .
calculate the linear width of central maximum on a screen kept 3 m away from the slit?
Here  = 600 nm = 600X10-9 m
Slit width a = 3 mm = 3 X10-3 m
D = 3 m.
Width of the central maximum 2x=2D/a
2x = 2D/a
=2X3X6X10-7/3 X10-3=1.2 mm
11. Determine the angular spread between central maximum and first order maximum of
diffraction pattern due to single slit of width 0.25 mm, when light of wave length 5890 A 0 is
incident on it normally.
Half the angular spread  is given by
A sin  = (2n +1)/2
If n = 1 , a sin  = 3 /2
When  is very small sin  =  
a  = 3 /2
 = 3 /2a = 3X5890X10-10/2X0.25X10-3
=3.534X 10-3 rad
Total angular speed is
=3.534X 10-3 rad
12. The R.I of a medium id 3. what is the angle of refraction if the unpolarized light in incident
on it at the polarizing angle of the medium?
 = 3
Tan ip =  = 3
Ip = tan-13
=600
As r = 900-ip
=900-600
R = 300
13. The spectral line of =65000 A in the light coming from a distant star is observed at 6525 0 .
determine the velocity of the star relative to earth?
=65000= 6500X10-10 m, ∆=25 A0
∆/=v/c
V=∆/Xc
=25/6500X3X108
=1.15X106 ms-1
HOT’S
1. for what distance is ray optics a good approximation when the aperture is 3 mm wide and
wave length is 500nm?
Here a= 3 mm = 3X10-3 m
 = 500 nm = 5X10-7 m
Tresnel distance Zf = a2/ = (3X10-3)/5X10-7
= 18 m
2. Two polarizing sheets are placed with their planes parallel, so that light intensity transmitted
is miximum through what angle must either sheet be transmitted so that light intensity drops
to half the maximum value?
According to malus law
I=I0 Cos2
Cos2 = I/I0= ½
Cos  = 1/2
 = 450 or  1350
The effect will be same when any of the two sheets turned through  in any direction.
3. Light reflected from surface from glass plate of refractive index 1.57 is linearly polarized
calculate the angle of refraction in glass?
N=1.57, r=?
According to Brewster’s law
Tan Ip = n = 1.57
Ip = tan -1(1.57)
= 57.50
R = 90-Ip = 32.50
4. A slit of 4 cm wide is irradiated with micro waves of wave length 2 cm. find the angular
spread of central maximum assuming incidence normal to the plane of the slit?
Here a = 4 cm = 4X10-2 m
 = 2 cm = 2 X10- 2 m
Angular spread of central maximum 2 is calculated from Sin=/a
= 2X10-2/4X10-2=1/2
= 300 then 2 = 2X300 = 600
5. A screen is placed 2 m away from the away from the single slit. Calculate the slit width if the
first minimum lies 5 mm on either side of the central maximum incident plane wave have a
maximum wave length of 5000A0.
Here the distance of the screen from the slit D = 2 m
X =5 mm =- 5X10-3 m
= 5000A0
For the first secondary minima Sin = /a = x/D
Then a = D/x = 2X5000X10-10/5X10-3
A=2X10-4 m
6.two spectral lines of sodium D1 and D2 have wave length of approximately 5890A0 and 5896 A0. a
sodium lamp sends incident plane wave on to a slit of width 2micrometer. A screen is located two
meters from the slit find the spacing between the first maxima of two sodium lines as measured on
the screen?
1 = 5890 A0 = 5890X10-10 m then a = 2X10-6 m
2 = 5896 A0 = 5896X10-10 m
D=2m
For the secondary maxima Sin= 3/21/a = x1/D
X1 = 31D/2a and x2 =32 D/2a
Spacing between the first secondary maxima of two sodium lines
= x2-x1=3D/2a(2-1)
= 3X2X6X10-10/2X2X10-6
9X10-4 m
7.
An unpolarized beam of light is incident on a group of four polarizing sheets which are
arranged in such a way that the characteristic direction of each polarizing sheet makes an
angle of 300 with that of the preceding sheet what fraction of incident un polarized light
incident?
If I0 is intensity of unpolarized light , then intensity of light from first polarized sheet is I0/2
Intensity of light from second polarized sheet I1 = I0/2(Cos 300)2 = I0/2(3/2)2
=I0/2 X3/4 = 3I0/8
Intensity of light from third polarized sheet I11=I1 Cos 300)2
I11=I0/2(3/4)X(3/2)2
Intensity of light from fourth polarized sheet I111 = I11X (Cos 300)2
I111= I0/2(3/4)2X(3/2)2
=I0X27/128
Therefore I111/I0=27/128.
8.
In diffraction of a single slit, a screen is placed 2m away from lens to obtain pattern. If the slit
width is 0.2mm and the first minimum lies 5mmon either side of central maximum, find the
wave length of light used.
D= 2m, d= 0.2 mm = 2X10-4 m
Then x1= 5 mm = 5X10-3 m
And x = D/d
=xd/D = 5X10-3X2X10-4/2 = 5X10-7 m
9
A plane wavefront of wave length 6X10-7 m falls on a slit 0.4 mm wide. A convex lens of focal
length 0.8 m placed behind the slit focuses the light on a screen . what is the linear diameter of
i) first minimum ii) second maximum?
Here =6X10-7 m ,a = 4X10-4 m and f = 0.8 m
a. linear diameter of first minimum 2x=2f/a = 2X10-7X0.8/0.4X10-3
= 2.4X10-3 m = 2.4 mm
b. Linear diameter of second maximum
2x1=2(2n+1)f/2a = 2X5X6X0.8X10-7/8X10-4
=6 mm
10 beam of light of wave length 600 nm from a distant source falls on a single slit 1mm wide and
the resulting diffraction pattern is observed on a screen two meters away . what is the
distance between first dark band on the either side of the central bright band?
Here dSin = n , n=1
dSin = 
d(x/D) = 
x = D/d
= 600X10-9X2/10-3
=12X10-4 m
EDITED BY
1.Mr. S.K.BHAT (VICE PRINCIPAL)
2. Mr. A.K.SHARMA (P.G.T.)
.
Wave Optics
Question 10.1:
Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the
wavelength, frequency and speed of (a) reflected, and (b) refracted light? Refractive index of water is
1.33.

Answer
Wavelength of incident monochromatic light,
λ = 589 nm = 589 × 10−9 m
Speed of light in air, c = 3 × 108 m/s
Refractive index of water, μ = 1.33
(a) The ray will reflect back in the same medium as that of incident ray. Hence, the wavelength,
speed, and frequency of the reflected ray will be the same as that of the incident ray.
Frequency of light is given by the relation,
Hence, the speed, frequency, and wavelength of the reflected light are 3 × 10 8 m/s, 5.09 ×1014 Hz,
and 589 nm respectively.
(b) Frequency of light does not depend on the property of the medium in which it is travelling. Hence,
the frequency of the refracted ray in water will be equal to the frequency of the incident or reflected
light in air.
Refracted frequency, ν = 5.09 ×1014 Hz
Speed of light in water is related to the refractive index of water as:
Wavelength of light in water is given by the relation,
Hence, the speed, frequency, and wavelength of refracted light are 2.26 ×108 m/s, 444.01nm, and
5.09 × 1014 Hz respectively.
Question 10.2:
What is the shape of the wavefront in each of the following cases:
(a) Light diverging from a point source.
(b) Light emerging out of a convex lens when a point source is placed at its focus.
(c) The portion of the wavefront of light from a distant star intercepted by the Earth.

Answer
(a) The shape of the wavefront in case of a light diverging from a point source is spherical. The
wavefront emanating from a point source is shown in the given figure.
(b) The shape of the wavefront in case of a light emerging out of a convex lens when a point source is
placed at its focus is a parallel grid. This is shown in the given figure.
(c) The portion of the wavefront of light from a distant star intercepted by the Earth is a plane.
Question 10.3:
(a) The refractive index of glass is 1.5. What is the speed of light in glass? Speed of light in vacuum is
3.0 × 108 m s−1)
(b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours
red and violet travels slower in a glass prism?

Answer
(a) Refractive index of glass, μ = 1.5
Speed of light, c = 3 × 108 m/s
Speed of light in glass is given by the relation,
Hence, the speed of light in glass is 2 × 108 m/s.
(b) The speed of light in glass is not independent of the colour of light.
The refractive index of a violet component of white light is greater than the refractive index of a red
component. Hence, the speed of violet light is less than the speed of red light in glass. Hence, violet
light travels slower than red light in a glass prism
Question 10.4:
In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4
m away. The distance between the central bright fringe and the fourth bright fringe is measured to be
1.2 cm. Determine the wavelength of light used in the experiment.

Answer
Distance between the slits, d = 0.28 mm = 0.28 × 10−3 m
Distance between the slits and the screen, D = 1.4 m
Distance between the central fringe and the fourth (n = 4) fringe,
u = 1.2 cm = 1.2 × 10−2 m
In case of a constructive interference, we have the relation for the distance between the two fringes
as:
Where,
n = Order of fringes = 4
λ = Wavelength of light used
∴
Hence, the wavelength of the light is 600 nm.
Question 10.5:
In Young’s double-slit experiment using monochromatic light of wavelengthλ, the intensity of light at a
point on the screen where path difference is λ, is K units. What is the intensity of light at a point
where path difference is λ /3?

Answer
Let I1 and I2 be the intensity of the two light waves. Their resultant intensities can be obtained as:
Where,
= Phase difference between the two waves
For monochromatic light waves,
Phase difference =
Since path difference = λ,
Phase difference,
Given,
I’ = K
When path difference
,
Phase difference,
Hence, resultant intensity,
Using equation (1), we can write:
Hence, the intensity of light at a point where the path difference is
is
units.
Question 10.6:
A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference
fringes in a Young’s double-slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for
wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the
wavelengths coincide?

Answer
Wavelength of the light beam,
Wavelength of another light beam,
Distance of the slits from the screen = D
Distance between the two slits = d
(a) Distance of the nth bright fringe on the screen from the central maximum is given by the relation,
(b) Let the nth bright fringe due to wavelength
and (n − 1)th bright fringe due to wavelength
coincide on the screen. We can equate the conditions for bright fringes as:
Hence, the least distance from the central maximum can be obtained by the relation:
Note: The value of d and D are not given in the question.
Question 10.7:
In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m
away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the
entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3.

Answer
Distance of the screen from the slits, D = 1 m
Wavelength of light used,
Angular width of the fringe in
Angular width of the fringe in water =
Refractive index of water,
Refractive index is related to angular width as:
Therefore, the angular width of the fringe in water will reduce to 0.15°.
Question 10.8:
What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)

Answer
Refractive index of glass,
Brewster angle = θ
Brewster angle is related to refractive index as:
Therefore, the Brewster angle for air to glass transition is 56.31°.
Question 10.9:
Light of wavelength 5000 Å falls on a plane reflecting surface. What are the wavelength and frequency
of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?

Answer
Wavelength of incident light, λ = 5000 Å = 5000 × 10−10 m
Speed of light, c = 3 × 108 m
Frequency of incident light is given by the relation,
The wavelength and frequency of incident light is the same as that of reflected ray. Hence, the
wavelength of reflected light is 5000 Å and its frequency is 6 × 1014 Hz.
When reflected ray is normal to incident ray, the sum of the angle of incidence,
reflection,
and angle of
is 90°.
According to the law of reflection, the angle of incidence is always equal to the angle of reflection.
Hence, we can write the sum as:
Therefore, the angle of incidence for the given condition is 45°.
Question 10.10:
Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and
wavelength 400 nm.

Answer
Fresnel’s distance (ZF) is the distance for which the ray optics is a good approximation. It is given by
the relation,
Where,
Aperture width, a = 4 mm = 4 ×10−3 m
Wavelength of light, λ = 400 nm = 400 × 10−9 m
Therefore, the distance for which the ray optics is a good approximation is 40 m.
Question 10.11:
The 6563 Å
line emitted by hydrogen in a star is found to be red shifted by 15 Å. Estimate the
speed with which the star is receding from the Earth.

Answer
Wavelength of
λ = 6563 Å
line emitted by hydrogen,
= 6563 × 10−10 m.
Star’s red-shift,
Speed of light,
Let the velocity of the star receding away from the Earth be v.
The red shift is related with velocity as:
Therefore, the speed with which the star is receding away from the Earth is 6.87 × 10 5 m/s.
Question 10.12:
Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than
the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed
of light in water? If not, which alternative picture of light is consistent with experiment?

Answer
No; Wave theory
Newton’s corpuscular theory of light states that when light corpuscles strike the interface of two media
from a rarer (air) to a denser (water) medium, the particles experience forces of attraction normal to
the surface. Hence, the normal component of velocity increases while the component along the
surface remains unchanged.
Hence, we can write the expression:
… (i)
Where,
i = Angle of incidence
r = Angle of reflection
c = Velocity of light in air
v = Velocity of light in water
We have the relation for relative refractive index of water with respect to air as:
Hence, equation (i) reduces to
But,
>1
Hence, it can be inferred from equation (ii) that v > c. This is not possible since this prediction is
opposite to the experimental results of c > v.
The wave picture of light is consistent with the experimental results.
Question 10.14:
Let us list some of the factors, which could possibly influence the speed of wave propagation:
(i) Nature of the source.
(ii) Direction of propagation.
(iii) Motion of the source and/or observer.
(iv) Wave length.
(v) Intensity of the wave.
On which of these factors, if any, does
(a) The speed of light in vacuum,
(b) The speed of light in a medium (say, glass or water), depend?

Answer
(a) Thespeed of light in a vacuum i.e., 3 × 108 m/s (approximately) is a universal constant. It is not
affected by the motion of the source, the observer, or both. Hence, the given factor does not affect
the speed of light in a vacuum.
(b) Out of the listed factors, the speed of light in a medium depends on the wavelength of light in that
medium.
Question 10.16:
In double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a
distant screen is 0.1º. What is the spacing between the two slits?

Answer
Wavelength of light used, λ = 6000 nm = 600 × 10−9 m
Angular width of fringe,
Angular width of a fringe is related to slit spacing (d) as:
Therefore, the spacing between the slits is
.
Question 10.17:
Answer the following questions:
(a) In a single slit diffraction experiment, the width of the slit is made double the original width. How
does this affect the size and intensity of the central diffraction band?
(b) In what way is diffraction from each slit related to the interference pattern in a double-slit
experiment?
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is
seen at the centre of the shadow of the obstacle. Explain why?
(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound
waves can bend around obstacles, how is it that the students are unable to see each other even
though they can converse easily.
(e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects
(observed when light propagates through small apertures/slits or around small obstacles) disprove this
assumption. Yet the ray optics assumption is so commonly used in understanding location and several
other properties of images in optical instruments. What is the justification?

Answer
(a) In a single slit diffraction experiment, if the width of the slit is made double the original width,
then the size of the central diffraction band reduces to half and the intensity of the central diffraction
band increases up to four times.
(b) The interference pattern in a double-slit experiment is modulated by diffraction from each slit. The
pattern is the result of the interference of the diffracted wave from each slit.
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is
seen at the centre of the shadow of the obstacle. This is because light waves are diffracted from the
edge of the circular obstacle, which interferes constructively at the centre of the shadow. This
constructive interference produces a bright spot.
(d) Bending of waves by obstacles by a large angle is possible when the size of the obstacle is
comparable to the wavelength of the waves.
On the one hand, the wavelength of the light waves is too small in comparison to the size of the
obstacle. Thus, the diffraction angle will be very small. Hence, the students are unable to see each
other. On the other hand, the size of the wall is comparable to the wavelength of the sound waves.
Thus, the bending of the waves takes place at a large angle. Hence, the students are able to hear
each other.
(e) The justification is that in ordinary optical instruments, the size of the aperture involved is much
larger than the wavelength of the light used.
Question 10.19:
A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction
pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5
mm from the centre of the screen. Find the width of the slit.

Answer
Wavelength of light beam, λ = 500 nm = 500 × 10−9 m
Distance of the screen from the slit, D = 1 m
For first minima, n = 1
Distance between the slits = d
Distance of the first minimum from the centre of the screen can be obtained as:
x = 2.5 mm = 2.5 × 10−3 m
It is related to the order of minima as:
Therefore, the width of the slits is 0.2 mm.
Question 10.20:
Answer the following questions:
(a) When a low flying aircraft passes overhead, we sometimes notice
a slight shaking of the picture on our TV screen. Suggest a possible explanation.
(b) As you have learnt in the text, the principle of linear superposition of wave displacement is basic
to understanding intensity distributions in diffraction and interference patterns. What is the
justification of this principle?

Answer
(a) Weak radar signals sent by a low flying aircraft can interfere with the TV signals received by the
antenna. As a result, the TV signals may get distorted. Hence, when a low flying aircraft passes
overhead, we sometimes notice a slight shaking of the picture on our TV screen.
(b) The principle of linear superposition of wave displacement is essential to our understanding of
intensity distributions and interference patterns. This is because superposition follows from the linear
character of a differential equation that governs wave motion. If y1 and y2 are the solutions of the
second order wave equation, then any linear combination of y1 and y2 will also be the solution of the
wave equation.
WAVE OPTICS
Very short short answer type question.
1.
2.
Define the term wave front? ‘
Define plane polarised light.
3.
4.
What is the value of refractive index of a medium of polarising angle 600?
What is the polarising angle of a medium of refractive index 1.732.
5.
6.
What is the polarising angle of a medium of refractive index √3.
A Plane wave front is incident noramally on a convex lens. Sketch the
refracted wave front.
7.
What type of wave front will imerge from (i) a point source and (ii) a
distant source of light.
8.
Draw a graf showing the variation of intensity of polarised light
Transmitted by an analyser.
9.
Name the phenomenon confirm the wave nature of light?
10.
State principle of superposition?
Short answer type question
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
What is the main condition to produce interferace of light?
Write two difference between interference and defraction.
Draw the pattern of wave front when a plane wave front is incident on a
prism and a convex lens.
Twho polaroid p1 and p2 are placed at 900 to each other. Find the
transmitted intesity if a third polaroid p3 is placed between p1 and p2 bisecting the angle
between them.
What is ment by coherent souce of light? Can two identical and
independent sodium lamp act as a coherent sources? Give reason for your answer.
Name the type of wave front that will be associated with (i) beam of
parallel rays and (ii) rays of light obtained by a linear source illuminated by another source
behind it.
What are coherent source of light? Draw the variation of intensity with
position in the interference pattern of young double slit experiment.
State two condition for sustained interference?
In a single slit diffraction experiment if the width of the slit is double,
how does the (i) intensity of light and(ii) width of central maximum change give reason for your
answer.
What is the affect on interference fringe in young double slit experiment
if (i) the separation between the slight is halved and (ii). The source is moved closer to the slit.
Justify your answer.
Long answer type question:
1.
2.
3.
4.
5.
How is a wave front different from a ray? Draw the geometrical wave
front when (i) light diverge from a point source (ii) light immeges out of a convex lens when a
point source is placed at its focus. State Huygenses principale with the help of suitable diagram
prove Snells law of refraction using Huygenes.
State Huygenes Priciple using the geometrical constructions of secondary
wave lens. Explain the refraction of plain wave front incident at plane surface. Hence verfy
snells law of refraction. Elustrate with the help of diagram the action convex lens and concave
mirror when plane wave front incident on it.
What is wave front. State the postulates of Huygenes. Using Huygenes
principle verify law of reflection of light.
What are coherent source of light. Obtain the condition for getting dark
and bright fringes in Youngs double slit experiment.
In Youngs double slit experiment deduce the condition for constructive
and destructive interference at a point source on the screen. Draw a graph showing the
6.
7.
8.
9.
10.
variation of the resultant intensity in the interference pattern against the position on the
screen.
What is interference of light? Write two essential condition for sustained
interference pattern to be produced on the screen. Draw a graph showing the variation of
intensity verses the position on the screen in Youngs experiment when (i) Both the slits are
open (ii) one of the slit is closed.
What is the affect on the interference pattern in Youngs double slit experiment when (i)
screen is more closer to the slit (ii) separtion between the slit is increased. Explain your answer.
What is diffraction of light? Draw a graph showing the variation of
intencity with angle ina single slit diffraction experiment. Write one feature which distingushes
the observed pattern from the double slit interference pattern. How would the diffraction
pattern of single slit the affected when (i) the width of the slit is decreased (ii) the
monocrometic source of light is replaced by white light.
What is menat by linearly polarised light? Which type of wave can be
polarised? Briefly explain a method for producing polarised light. Two poloroid are placed to 900
to each other and the intensity of transmitted light is zero. What will be the intensity of
transmitted light when one more polaroid is placed between these two bisecting the angle
between them. Take the intensity of unpolarised light as I0
What are polaroid? How are they used to deomonsarate (i) light waves
are transverse in nature. (ii) If an unpolarised light is incident then light wave will get linearly
polarised? What is Brewsters angle? When an upolarised light is incident on a plane glass
surface what should be the angle of incident so that the reflected and refracted rays are
perpendicular to each other.
A parrallel beam of monochromatic light falls normally on narrow slit and
the light coming out of the slit is obtained on screen kept behind parrallel to the slit plane. What
kind of pattern do we observe on the screen and why? How does the angular width and linerar
width of the Principal maximum in thi pattern change when the screen is more parallel to itself.
State two point of difference between this pattern and the interference pattern observe in the
Young double slit experiment.