Physics II

Using a calculator in degrees mode and giving an argument to a trigonometric

function in radians or vice versa will often result in mistakes.
 
Two or more waves are said to be A monochromatic source, like a laser,
coherent if they have the same will emit light of one specific
frequency and, therefore, a constant wavelength.
phase difference from each other. A polychromatic source, like any
white light source, will emit multiple
wavelengths.


Young's double
Diffraction
slit

Refraction Fresnel's biprism

Interference
Lloyd's mirror

Reflection Thin films Newton's rings

Michelson
interferometer

Light sources must be (a)coherent, and (b)have the same amplitude and
frequency.
Light sources should be (c) monochromatic.

We shine a monochromatic light of wavelength (λ) on two slits separated
by a distance (d) then view the interference pattern on a screen at a
distance (L) from the slits.

A pattern of alternating bright and dark fringes each of order (m) appears
where the central fringe is bright of order zero (m=0).

We observe the distance of each fringe from the center (y), and the angle
the rays reaching a fringe makes with the center line (θ). We can calculate
the phase difference (ϕ), and the path difference (δ) between the rays
that interfered (superposition) to produce the fringe.

The distance between two consecutive similar fringes is (Δy) (fringe

width).
 λ: wavelength of the monochromatic light used.
d: slits separation.
L: distance between the slits and the screen.
m: order of a fringe.
y: distance between center of the screen and a fringe.
θ: angle between the fringe line and the central line.
ϕ: phase difference between the rays that form the fringe.
δ: path difference between the rays that form the fringe.
∆y: fringe width (distance between a point and the next point of the same
intensity).
Bright = constructive = strong = maximum intensity
Dark = Destructive = weak = minimum intensity

second second first first Central first first second second

bright dark bright dark bright dark bright dark bright
m 2 1 1 0 0 0 1 1 2
y 2𝜆𝑙 1.5𝜆𝑙 𝜆𝑙 0.5𝜆𝑙 0 0.5𝜆𝑙 𝜆𝑙 1.5𝜆𝑙 2𝜆𝑙
𝑑 𝑑 𝑑 𝑑 𝑑 𝑑 𝑑 𝑑
ϕ 4𝜋 𝑟𝑎𝑑 3𝜋 𝑟𝑎𝑑 2𝜋 𝑟𝑎𝑑 1𝜋 𝑟𝑎𝑑 0 1𝜋 𝑟𝑎𝑑 2𝜋 𝑟𝑎𝑑 3𝜋 𝑟𝑎𝑑 4𝜋 𝑟𝑎𝑑
δ 2𝜆 1.5𝜆 𝜆 0.5𝜆 0 0.5𝜆 𝜆 1.5𝜆 2𝜆
 
Reflected light will experience a 180° or π radians Reflected light will experience no phase change (a
phase change (a hard reflection) when it reflects soft reflection) when the reflection is from a
from a medium with a greater index of refraction. medium with a smaller index of refraction.

Φ=180° Φ=0°

t: thin film thickness

n: thin film’s material refractive index. Φ1
Φ2
λ0: wavelength of light in free space.

(thin film problems have multiple solutions

for n, t, λ0 at different values of m)

Phase relation Φ1 = Φ2 Φ1 ≠ Φ2
Interference type
constructive
m= 1,2,3…etc. m= 0,1,2,3…etc.
destructive
m= 0,1,2,3…etc. m= 1,2,3…etc.

Central fringe is dark.

A special case of thin film where Φ1 ≠ Φ2

Where (R) is the radius of curvature, (r) is the distance from the center of
the ring to the fringe under consideration (radius of the fringe), and (t) is
the thickness of the thin film at the point under consideration.

Note: n is the refractive index of the thin film between the two pieces of
glass not the refractive index of the glass.
The thickness (t) starts at zero at the center of the apparatus and
increases as we move away from the center.
Constructive (bright) Destructive (dark)

2𝑛𝑡 = (𝑚 + 0.5) 𝜆 2𝑛𝑡 = 𝑚𝜆

r r
t= t=
2𝑅 2𝑅
(𝑚 + 0.5)𝜆 𝑅 𝑚𝜆 𝑅
r = r =
𝑛 𝑛
central dark fringe: m=0
First dark fringe: m=1
First bright fringe: m = 0


If mirror M1 is moved a distance (d) the pattern alternates between

bright and dark (m) times according to the relation.
2d=mλ


The same pattern as young’s double slit but uses a biprism instead of the
double slit.

Similar pattern to young’s double slit but the central fringe is dark, and
the setup uses a mirror where the two coherent sources are the actual
source (s) and its virtual image (s′) that produced by reflection from the
mirror.


Using Young’s double slit and or Fresnel’s biprism to measure the
wavelength (λ) of light by counting the number of a single type of
fringes (m) contained in a distance (h) from the center.
𝒉 𝒅
𝛌= ×
𝒎 𝒍

𝑖𝑠 𝑡ℎ𝑒 𝑓𝑟𝑖𝑛𝑔𝑒 𝑤𝑖𝑑𝑡ℎ (∆𝑦).

𝑚
𝑖𝑠 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑟𝑖𝑛𝑔𝑒𝑠 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑙𝑒𝑛𝑔𝑡ℎ.
ℎ
Calculating the thickness (t) and refractive index (n) of a thin film
using Young’s double slit or Fresnel’s biprism by observing the change
of the position of the central bright fringe when a thin film is added to
the path of one of the slits.
𝐝𝐲′
𝛅 = (𝐧 − 𝟏)𝐭 = 𝐦𝛌 = 𝐝𝐬𝐢𝐧(𝜽) ≈
𝐋
where (m) is the number of fringes shifted due to the thin film.

Calculating the thickness (t) and refractive index (n) of a thin film or
the refractive index (n) of a gas using Michelson interferometer by
observing the change of the number of fringes shifted when a thin film
is introduced to the path of one of a ray.
𝟐(𝐧 − 𝟏)𝐭 = 𝐦𝛌
 (I0 is the maximum light
intensity at the center)

(ϕ in radians)

(at small θ)

( ) ( )
 Phase relation Φ1 = Φ2 Φ1 ≠ Φ2
Interference type
constructive
m= 1,2,3…etc. m= 0,1,2,3…etc.
destructive
m= 0,1,2,3…etc. m= 1,2,3…etc.

Constructive (bright) Destructive (dark)

2𝑛𝑡 = (𝑚 + 0.5) 𝜆 2𝑛𝑡 = 𝑚𝜆

r r
t= t=
2𝑅 2𝑅
(𝑚 + 0.5)𝜆 𝑅 𝑚𝜆 𝑅
r = r =
𝑛 𝑛

Measuring wavelength using Young’s 𝒉 𝒅

𝛌= ×
double slit experiment. 𝒎 𝑳
Calculating thin film properties using 𝐝𝐲′
𝛅 = (𝐧 − 𝟏)𝐭 = 𝐦𝛌 = 𝐝𝐬𝐢𝐧(𝜽) ≈
Young’s double slit. 𝐋
Calculating thin film properties using 𝟐(𝐧 − 𝟏)𝐭 = 𝐦𝛌
Michelson interferometer.
Example 1: In Young’s double slit experiment it’s found that blue light of
wavelength 440 nm gives a second order maximum at a certain location on the
screen. What wavelength of visible light would have the minimum at the same
location?

𝑦 =𝑦
𝑚𝜆 𝐿 (𝑚 + 0.5)𝜆 𝐿
=
𝑑 𝑑
2 × 440 × 10 = (𝑚 + 0.5)𝜆
For m = 0 𝜆 = 1760 𝑛𝑚 invisible light.
For m = 1 𝜆 = 𝟓𝟖𝟔. 𝟔𝟕 𝒏𝒎 visible light.
For m = 2 𝜆 = 352 𝑛𝑚 invisible light.

Example 2: In Newton’s rings experiment, the radius of the 12th, dark ring
changes from 1.4 cm to 1.27 cm when a liquid is introduced between the lens
and the plate. Calculate the refractive index of the liquid.

𝑚𝜆 𝑅
r =
𝑛

1.4 × 10 = 12 × 𝜆 × 𝑅 (1)

𝑚𝜆 𝑅
r =
𝑛
 12 × 𝜆 × 𝑅
1.27 × 10 = (2)
𝑛

Dividing (1) by (2)

1.4
= √𝑛
1.27
∴ 𝑛 ≈ 1.215

Example 3: In Young's double slit experiment a distance h=10 mm on the

screen contains 3 fringes. If the slit separation is 0.25 mm and the distance
between the slits and the screen is 2 m (a) determine the wavelength and the
fringe width. (b) If the experiment is done under water (n=1.33) determine
the wavelength in water and the fringe width. (c) how many fringes are
contained in distance 10 mm under water

(a)
ℎ 𝑑 10 × 10 × 0.25 × 10
λ = × = = 416.7 𝑛𝑚
𝑚 𝐿 3×2
ℎ 10 × 10
∆𝑦 = = = 3.33 𝑚𝑚
𝑚 3
(b)
λ 416.7 × 10
λ = = = 313.28 𝑛𝑚
𝑛 1.33
λ 𝐿 313.28 × 10 × 2
∆𝑦 = = = 2.506 𝑚𝑚
𝑑 0.25 × 10
(c)
ℎ 10 𝑚𝑚
𝑚= = = 4 𝑓𝑟𝑖𝑛𝑔𝑒𝑠
∆𝑦 2.506 𝑚𝑚
1- A pair of narrow, parallel slits separated by 0.25 mm is illuminated by
green light (λ = 546.1 nm). The interference pattern is observed on a
screen 1.2 m away from the plane of the slits. Calculate the distance (a)
from the central maximum to the first bright region on either side of the
central maximum and (b) between the first and second dark bands.

× . × × .
(a) 𝑦 ( ) = = = 2.621 𝑚𝑚
. ×
. × × .
(b) ∆𝑦 = = = 2.621 𝑚𝑚
. ×

2- A laser beam (λ = 632.8 nm) is incident on two slits 0.2 mm apart.

Approximately how far apart are the bright interference lines on a screen
5.0 m away from the double slits?

𝜆𝐿 632.8 × 10 × 5
∆𝑦 = = = 15.82 𝑚𝑚
𝑑 0.2 × 10

3- Young's double-slit experiment is performed with 589 nm light and a slit-

to-screen distance of 2 m. The tenth interference minimum is observed
7.26 mm from the central maximum. Determine the spacing of the slits.

10th dark means m = 9

( . ) . × × ×
𝑑= = =1.541 mm
. ×
4- Two narrow parallel slits separated by 0.85 mm are illuminated by λ =
600 nm light, and the viewing screen is 2.8 m away from the slits.
(a) What is the phase difference between the two interfering waves on a
screen at a point 2.5 mm from the central bright fringe.
(b) What is the ratio of intensity at this point (y=2.5 mm)the center of a
bright fringe?

ϕ
× . × × . × 𝜙= δ
(a) 𝜙 = × = ≈ 7.947 𝑟𝑎𝑑
× × .
. . δ≈
(b) = cos = cos ≈ 0.453 y δ

5- A thin film of oil (n= 1.25) covers a smooth wet pavement (n water=1.33).
When viewed perpendicular to the pavement, the film appears to be
predominantly red (λ = 640 nm ) and has no blue (λ = 512 nm ). How
thick is it?

Φ1= Φ2
Constructive (red) Destructive (blue)
𝑚𝜆 (𝑚 + 0.5)𝜆
𝑡= 𝑡=
2𝑛 2𝑛
𝑡 = 256 𝑛𝑚 𝑡 = 102.4 𝑛𝑚
𝑡 = 512 𝑛𝑚 𝑡 = 307.2 𝑛𝑚
𝑡 = 768 𝑛𝑚 𝑡 = 512 𝑛𝑚
∴ 𝑡 = 512 𝑛𝑚
6- Mirror M1 is displaced a distance L in Michelson interferometer, during
this displacement, 250 fringes ( formation of successive dark or bright
bands) are counted . The light being used has a wavelength of 632.8 nm.
Calculate the displacement L .

∆L = = 79.1 𝜇𝑚

7- Calculate the minimum thickness of a soap bubble film ( n= 1.33) that

results in constructive interference in the reflected light if the film is
illuminated with light whose wavelength in free space is 600 nm. What
other film thicknesses produce constructive interference?

Φ1 ≠ Φ2

(𝑚 + 0.5)𝜆
𝑡=
2𝑛
m= 0,1,2,3…etc.

( )
 MCQ

1. The main principle used in Interference is _____

a) Heisenberg’s Uncertainty Principle.
b) Superposition Principle.
c) Quantum Mechanics.
d) Fermi Principle.

2. When Two waves of same amplitude add constructively, the intensity

becomes _________
a) Double.
b) Half.
c) Four times
d) One-fourth.

3. The shape of the pattern depends on the ________

e) Distance between the slits.
f) Distance between the slits and the screen.
g) Wavelength of light.
h) Shape of the slit.

4. Which phenomenon is observed in the following figure?

a. Wedge-Shaped film
b. Destructive Interference
c. Refraction
d. Newton’s Rings
5. In Young's double-slit experiment, the slit separation is doubled. To
maintain the same fringe spacing on the screen, the screen-to-slit distance D
must be changed to:
a. D/2
b. 2D 𝜆𝐿 𝜆𝐷
∆𝑦 = =
c. 4D 𝑑 𝑑
d. D/2^0.5

1. For destructive interference, path difference is even number of half

wavelengths ( T / F )

(Explanation: destructive interference occurs at path differences of 0.5λ,

1.5λ , 2.5λ , 3.5λ …etc. i.e. at odd number of half wavelengths (1×0.5λ)
(3×0.5λ) (5×0.5λ) (7×0.5λ).

2. Interference is observed only when the phase difference between the two
waves is zero ( T / F )

(Explanation: interference occurs at any phase difference and is observed

for coherent light at any constant phase relation where it’s completely
constructive when the phase difference is an even multiple of half cycles
i.e. 0, 2π, 4π, 6π …etc. and is completely destructive when the phase
difference is an odd multiple of half cycles i.e. π, 3π, 5π …etc.)

3. In Newton’s ring experiment, the diameter of bright ring is directly

proportional to the wavelength ( T / F )

(Explanation: the diameter of bright ring is directly proportional to the

square root of the wavelength)
4. Michelson’s interferometer experiment consists of two acute angled
prisms placed base to base ( T / F )

(Explanation: Fresnel’s biprism consists of two acute angled prisms

placed base to base)

5. In a double slit experiment, the sources are placed 0.2 mm apart and the
fringes are observed on a screen 50 cm away. If the fourth bright fringe is
situated at a distance 12mm from the central fringe, the wavelength of
the light = 1.2 micrometer ( T / F )

(Explanation: λ = = 1.2 𝜇𝑚)