PY1001
Interference
Q1. What is the necessary condition on the path length difference (and
phase difference) between two waves that interfere (a) constructively and
(b) destructively ?
A1.
(a) Two waves interfere constructively if their path difference is zero, or an
integral multiple of the wavelength, according to δ=mλ, with m=0,1,2,3,…
(b) Two waves interfere destructively if their path difference is a half
wavelength, or an odd multiple of λ/2, described by:
δ=(m + 1/2)λ, with m = 1,2,3,…
Q2. Obtain an expression for the fringe-width in the case of interference
of light of wavelength λ, from a double-slit of slit-separation d.
A2.
Q3. Explain the term coherence.
A3.
For interference to occur, Phase difference at a point must not change wrt time.
This is possibly only when 2 sources are completely coherent.
2 waves are said to be coherent when they are of:
• Same amplitude
• Same frequency
• Same phase or a constant phase difference
Q4. Obtain an expression for the intensity of light in double-slit
interference using phasor-diagram.
A4.
INTENSITY IN DOUBLE SLIT INTERFERENCE
Resultant of E1= E0 sin ωt & E2= E0 sin (ωt + f)
Phasor à Rotating vector.
ADDITION OF TWO VECTORS USING PHASORS
E2
Let two vectors be, E1= E0 sin ωt &
E0
E0
E1
E2= E0 sin (ωt + f)
Resultant field E = E1 + E2
ωt + f
ωt
INTENSITY IN DOUBLE SLIT INTERFERENCE
Resultant of E1= E0 sin ωt & E2= E0 sin (ωt + f)
From phasor diagram,
E = E1 + E2
E0
= Eq sin(wt + b)
f
E2
Eq
= 2E0 cos b sin(wt + b)
But b = f/2. So above eqn can be
written as,
b
E
b
E0
E1
E = 2 E0 cos(f/2) sin(wt+f/2)
ωt
INTENSITY IN DOUBLE SLIT INTERFERENCE
§ E = 2 E0 cos(f/2) sin(wt+f/2)
§ So intensity at an arbitrary point P on the screen due to
interference of two sources having phase difference f;
I
µ
æfö
4 E02 cos 2 ç ÷
è2ø
æfö
4 I 0 cos 2 ç ÷
è2ø
where I = E2 is intensity due to single source
0
0
I
µ
Q5. Draw a schematic plot of the intensity of light in a double-slit
interference against phase-difference (and path-difference).
A5.
Q6. Explain the term reflection phase-shift.
A6.
It has been observed that if the medium beyond the interface has a higher index of
refraction, the reflected wave undergoes a phase change of pi (=180o).
If the medium beyond the interface has a lower index of refraction, there is no phase
change of the reflected wave.
Phase changes on reflection at a junction
between two strings of different linear mass
densities.
Q7. Obtain the equations for thin-film interference.
A7.
INTERFERENCE FROM THIN FILMS
Equations for Thin Film Interference:
Normal incidence (qi = 0)
Path difference = 2 d + (½) ln
­
BACK SURFACE
Constructive interference:
2 d + (½) ln = m ln
m = 1, 2, 3, . . .
(maxima)
Destructive interference:
2 d + (½) ln = (m+½) ln
m = 0, 1, 2, . . .
(minima)
Q8. Explain the interference-pattern in the case of wedge-shaped thinfilms.
A8.
INTERFERENCE FROM THIN FILMS
WEDGE SHAPED FILM
Q9.
an
Obtain
In wedge – shaped thin film,
constructive interference occurs in
certain part of the film [2 d + (½) ln =
m ln] and destructive interference in
others [2 d + (½) ln = (m+½) ln].
Then bands of maximum and
minimum intensity appear, called
fringes of constant thickness.
expression for the radius of mth order bright ring in the case of
Newton’s rings.
A9.
Q10. Explain Michelson’s interferometer. Explain how microscopic length
measurements are made in this.
A10.
(a)
• Light from an extended monochromatic source P falls on a half-silvered
mirror M.
• The incident beam is divided into re ected and transmitted beams of equal
intensity.
• These two beams travel almost in perpendicular directions and will be
re ected normally from movable mirror (M2) and xed mirror (M1).
• The two beams nally proceed towards a telescope (T) through which
interference pattern of circular fringes will be seen.
ff
fi
fl
fi
fl
• The interference occurs because the two light beams travel di erent paths
between M and M1 or M2.
• Each beam travels its respective path twice. When the beams recombine,
their path di erence is 2 (d2 – d1)
ff
(b)
The path difference can be changed by moving mirror M2. As M2 is moved, the
circular fringes appear to grow or shrink depending on the direction of motion of M2.
New rings appear at the center of the interference pattern and grow outward or larger
rings collapse disappear at the center as they shrink.
For the center of the fringe pattern to change from bright dark and to bright again, the
path difference between two beams must change by one wavelength, which means
that mirror M2 moves through a distance of λ/2. If N fringes cross the field of view
when mirror M2 is moved by △d, then
△d = N (λ/2)
△d is measured by a micrometer attached to M2. Thus microscopic length
measurements can be made by this interferometer.
Diffraction
Q1. Discuss the diffraction due to single-slit. Obtain the locations of the
minima and maxima qualitatively.
A1.
SINGLE-SLIT DIFFRACTION
At point P1,
path difference between r1
and r2 is
(a/2) sinq
So the condition for first minimum,
a
l
sin q =
2
2
or
a sin q = l
This is satisfied for every pair of rays, one of which is from upper half
of the slit and the other is a corresponding ray from lower half of the
slit.
SINGLE-SLIT DIFFRACTION
At point P2,
path difference between
r1 and r2 is (a/4) sinq
Q2.
So the condition for second minimum,
a
l
sin q =
or
a sin q = 2l
4
2
This is satisfied for every pair of rays, separated by a distance a/4.
In general, the condition for m TH minima,
a sin q = ml
m = ± 1, ± 2, ± 3, . . .
There is a secondary maximum approximately half way between
each adjacent pair of minima.
Obtain an expression for the intensity in single-slit diffraction pattern,
using phasor-diagram.
A2.
INTENSITY IN SINGLE – SLIT DIFFRACTION
From diagram,
Eq = 2 R sin
f
2
Em
R
Combining,
E
f
Eq = m sin
f
2
2
sin a
Or , Eq = Em
Also f =
where a =
f
2
a
INTENSITY IN SINGLE – SLIT DIFFRACTION
f is the phase difference
between rays from the top
and bottom of the slit.
INTENSITY
So we can
write, IN SINGLE – SLIT DIFFRACTION
2
The
2pintensity I q µ E q
f=
l
a sin q
=
2 æ sin a ö
Em
ç
÷
è a ø
2
2
æ sin a ö
2
If
mç
pa÷ø where I m µ Em is the max. intensity
a
è
So, a = =
sin q
2 above
l eqn., for minima, sina = 0
From the
Iq =
Þ a = m p where m = ±1,±2,±3,.....
or, a sin q = m l where m = ±1,±2,±3,.....
Q3.
Calculate, approximately, the relative intensities of the first three
secondary maxima in the single-slit diffraction pattern.
INTENSITY IN SINGLE – SLIT DIFFRACTION
A3.
The intensity distribution in
single-slit diffraction for three
different values of the ratio a/λ
The intensity distribution in
single-slit diffraction for three
different values of the ratio a/l
Q4. Discuss qualitatively diffraction at a circular aperture.
A4.
• The mathematical analysis at a circular aperture shows that the first minimum
occurs at an angle from the central axis given by:
sin(x) = 1.22λ/d ; d = Diameter of aperture
• The equation for the first minimum in a single slit diffraction is:
sin(x) = λ/a ; a = Slit Width
Q5. Explain Rayleigh’s criterion for resolving images due to a circular
aperture.
A5.
The image of two closely resolved sources is said to be resolved if the angular
separation of the 2 point sources is such that the central maxima of the
diffraction pattern of the source falls on the first minima of the diffraction
pattern of the other.
x = sin^-1(1.22λ/d)
Since, x is very small. Thus,
x = 1.22λ/d
x = Smallest angular separation for which we can resolve the images of two
objects.
Adding all the phasors, we get the resultant E1 due to the first slit.
x is the phase difference between the light waves at the point P,
emitted from bottom edge of the first slit and top edge of the
second slit. E2 is the resultant due to the second slit. Eq is the
resultant of E1 and E2.
DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
Each of the two slits is divided into N zones. Electric field at P is
found by adding the phasors. There is phase difference of Df =
f/N between each of the N phasors where f is the phase
difference between1st phasor and Nth phasor.
Q6. Obtain an expression for the intensity in double-slit diffraction
pattern,
using phasor-diagram.
DOUBLE-SLIT
INTERFERENCE AND DIFFRACTION COMBINED
From the figure,
d
E q = 2E1 sin
2
f
f
where + d + + x = p
2
2
or
d = p - ( x + f)
d
æp x+fö
æx+fö
= sinç ÷ = cosç
÷ .........( A )
2
2 ø
è2
è 2 ø
x p
and
= (d - a) sin q
2 l
f pa
Adding =
sin q to both sides of above eqn, we get,
2
l
x+f p
= d sin q which is b
2
l
Also
sin
A6.
DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COMBINED
Substituti ng this in eqn( A ), we get,
d
sin = cos b
2
From sin gle - slit diffraction, we have,
the electric amplitude at P due to one slit,
æ sin a ö
E1 = Em ç
÷
è a ø
d
\ E q = 2E1 sin
2
æ sin a ö
ie, E q = (2Em )ç
÷ cos b
è a ø
æ sin a ö
\ I q = I m (cos b) ç
÷
è a ø
2
2
DOUBLE-SLIT
INTERFERENCE PATTERN
**Q7.
SINGLE-SLIT DIFFRACTION
PATTERN
Discuss qualitatively the diffraction due to multiple slits (eg, 5
slits).
A7.
Multiple slit arrangement will be the
interference pattern multiplied by the single
slit diffraction envelope. This assumes that
all the slits are identical.
Condition for principal maxima,
d*sin(theta) = mλ
where d is the separation between adjacent
slits.
Location of principal maxima is
independent of number of slits.
Q8. Obtain an expression for the width of the central maximum in diffraction
pattern due to multiple slits.
A8.
MULTIPLE SLITS
Width of the maxima: Central maximum
§ The pattern contains central maximum with minima on
either side.
§ At the location of central maximum, the phase difference
between the waves from the adjacent slits is zero.
§ At minima, the phase difference is such that,
Df =
2p
where N is the number of slits
N
§ Corresponding path difference is,
l
æ l ö
DL = ç
÷Df =
N
è 2p ø
MULTIPLE SLITS
Width of the maxima: Central maximum
l
æ l ö
DL = ç
÷Df =
N
è 2p ø
§ Also we know,
DL = d sin dq 0
l
= d sin dq 0
N
l
sin dq 0 =
Nd
l
dq 0 »
Nd
From the equation, for given l and
d if we increase number of slits (N),
then the angular width of principal
maximum decreases. ie the
principal
maximum
becomes
sharper.
MULTIPLE SLITS
Width of the maxima: Other principal maxima
d sin(θ + dθ )
=
mλ +
λ
N
é
ù
d êsin q cos
dq
+
cos
q
sin
dq
#"!
#"! ú
êë
dq ú
1
û
d
#sin
"!q + (d cos q) dq
!
ml + (d cosq) Dq
=
=
= ml +
l
N
ml + l N
ml + l N
l
MULTIPLE
SLITS
ANGULAR HALF WIDTH OF mTH
dq =
PRINCIPAL MAXIMUM AT q
cos
q
Width N
of d
the
maxima:
Other principal maxima
The principal maximum become sharper as
number
(N) increases
th principal
For theofmslits
Q9.
maximum at q by a
grating: d sinq = m l.
For the first minimum
at q + dq after the mth
principal maximum
d sin(θ + dθ ) =
mλ +
MINIMUM AT θ
+dθ
mth PRINCIPAL
MAXIMUM AT θ
λ
N
MINIMUM AT θ
+dθ
mth PRINCIPAL
MAXIMUM AT θ
Obtain an expression for the width of a principal maximum at an angle
in diffraction pattern due to multiple slits.
A9.
Q10. Obtain an expression for dispersion by a diffraction grating.
A10.
D = △theta/△λ
& d*sin(theta) = mλ
On differentiating:
d*cos(theta)*△theta = mλ*△λ
Thus, D = △theta/△λ = m/d*cos(theta)
Q11. Discuss Bragg’s law for X-ray diffraction.
A11.
• In every crystal, several sets of parallel planes called the Bragg planes can be
identified.
• Each of these planes have an identical and a definite arrangement of atoms.
• Different sets of Bragg planes are oriented at different angles and are characterized
by different inter planar distances d.
Theta = Glancing Angle i.e. qangle between the incident x-ray beam and the
reflecting crystal planes.
For constructive interference of diffracted x-rays the path difference for the rays from
the adjacent planes, (abc in the figure) must be an integral number of wavelength.
2d*sin(theta) = nλ
Q12. Obtain an expression for resolving power of a diffraction grating.
A12.
For two close spectral lines of wavelength λ1 and λ2, just resolved by the grating, the
resolving power is defined as
R = λ/△λ ; △λ = l λ1 - λ2 l ; λ = (λ1 + λ2)/2
We have, D = △theta/△λ = m/d*cos(theta)
& △theta = λ/Ndcos(theta)
Thus, R = λ/△λ = Nm
Polarisation
Q1. Sketch the schematic graph of a travelling electromagnetic wave
showing the electric and magnetic vectors.
A1.
Q2. Explain the law of Malus with a diagram.
A2.
Malus's Law. According to Malus, when
completely plane polarized light is incident on the
analyzer, the intensity I of the light transmitted by
the analyzer is directly proportional to the square
of the cosine of angle between the transmission
axes of the analyzer and the polarizer.
I = I(max)*cos 2 x
Q3. Explain with diagram, the polarization of reflected light, incident at
Brewster’s angle.
A3.
• If light is incident on the surface of a
dielectric material at polarizing angle of
incidence, (Theta(p) = Brewster’s Angle) the
reflected light is completely plane polarized
and the transmitted ray is perpendicular to
reflected ray.
•
n2/n1 = tan(theta(p))
• This expression is called Brewster’s law, and
the polarizing angle theta(p) is sometimes
called Brewster’s angle.
• n varies with wavelength for a given substance, Brewster’s angle is also a
function of wavelength.
Q4. Explain the method of producing plane-polarized light by refraction
in a stack of glass plates.
A4. Unpolarized light is incident at the angle qp. All reflected lights are
polarized perpendicular to the plane of figure. After passing through the several
layers, the transmitted wave no longer contains any appreciable component
polarized perpendicular to the figure.
Q5. Explain the phenomenon of double refraction with a diagram
indicating the directions of polarizations for the two beams.
A5.
DOUBLE REFRACTION
Unpolarized light incident on a birefringent
material (eg. calcite crystal) splits into an
ordinary (o) ray and an extraordinary (e) ray.
These two rays are polarized in mutually
perpendicular directions.
The o-wave travels in the crystal with the same speed vo in all directions. The
o-ray obeys Snell’s Law of refraction. The crystal has a single index of
refraction no for o-wave.
The e-wave travels in the crystal with a speed that varies with direction from
vo to ve. It does not obey the Snell’s Law. The index of refraction of the crystal
varies with direction from no to ne for the e-wave.
Quantum Mechanics:
Q1. What is a wave function? What is its physical interpretation?
A1. The amplitude of the de Broglie wave associated with a particle is called
probability amplitude, or the wave function, and is denoted by Ψ.
Max Born proposed that the wave function 'F for a beam of particles be
interpreted in this same way, namely, that its square is a direct measure of the
average density of particles in the beam.
Q2. What are the mathematical features of a wave function?
A2. The important mathematical features of a physically acceptable wave
function Ψ(x) for a system are:
• Ψ(x) may be a complex function or a real function, depending on the system;
• Ψ(x), must be finite, continuous and single valued every where;
• The space derivatives of Ψ, must be finite, continuous and single valued
every where;
• Ψ must be normalisable.
Q3. By solving the Schrödinger equation, obtain the wave-functions for a
particle of mass m in a one-dimensional “box” of length L.
A3. Till slide 23
Q4. Apply the Schrödinger equation to a particle in a one-dimensional
“box” of length L and obtain the energy values of the particle.
A4. Till slide 23
Q5. Sketch the lowest three energy states, wave-functions, probability
densities for the particle in a one-dimensional “box”.
A5. Till slide 23
Q6,7: Register
Q8. Sketch the potential-well diagram of finite height U and length L,
obtain the general solution of the Schrödinger equation for a particle of
mass m in it.
A8. Till slide 31.
Q9. Sketch the lowest three energy states, wave-functions, probability
densities for the particle in a potential well of finite height.
A9. Slide 32
Q10. Give a brief account of tunnelling of a particle through a potential
energy barrier.
A10. Till slide 37
Q11. Give a brief account of the quantum treatment of a simple
harmonic oscillator.
A11.