Young’s double slit experiment

1. CALCULATION OF FRINGE WIDTH

This is a classic example of interference effects in light waves. Two light rays pass through two slits, separated by a
distance d and strike a screen a distance, D, from the slits, as in Fig. 1.

If 𝑑 ≪ 𝐷 then the difference in path length 𝑟1 − 𝑟2 travelled by the two rays is approximately: 𝑟1 − 𝑟2 ≈ 𝑑 sin 𝜃

Where 𝜽 is approximately equal to the angle that the rays make relative to a perpendicular line joining the slits to the
screen.

If the rays were in phase when they passed through the slits, then the condition for constructive interference at the
screen is: 𝒅 𝐬𝐢𝐧 𝜽 = m𝝀, 𝒎 = ± 1, ±2,...
P
Whereas the condition for destructive interference at the screen is:
𝟏
𝒅 𝐬𝐢𝐧 𝜽 = (m + )λ , 𝒎 = ± 1, ± 2,... A y
𝟐

The points of constructive interference will appear as bright bands 𝜽

𝑑 𝜽 O
and the points of destructive interference will appear as dark bands. D
B
These dark and bright spots are called interference fringes.

Note: In the case that 𝑦, the distance from the interference fringe to the

point of the screen opposite the centre of the slits is much less than
𝒚
𝐷 (𝑦 ≪ 𝐷), one can use the approximate formula: 𝐬𝐢𝐧 𝜽 ≈
𝑫

So that the formulas specifying the y - coordinates of the bright and dark spots, respectively are:
𝒎𝝀𝑫
𝒎=
𝒚𝑩 ………………………………..for bright spots
𝒅
𝟏
(𝒎+𝟐)𝝀𝑫
𝒎=
𝒚𝑫 ……………………………for dark spots
𝒅
𝝀𝑫
The spacing between the bright spots is: ∆y =𝑦𝑚+1
𝐵 𝐵
− 𝑦𝑚 =
𝒅
𝝀𝑫
& the spacing between the dark spots is: ∆y =𝑦𝑚+1
𝐷 𝐷
− 𝑦𝑚 =
𝒅
𝝀𝑫
It can be seen that the spacing of the fringes (fringe width ∆y=β= ) depends on the wavelength, the separation of the
𝒅
holes, and the distance between the slits and the observation plane, as noted by Young.

 If d < < D then the spacing between the interference can be large even when the wavelength of the light is

very small (as in the case of visible light). This give a method for (indirectly) measuring the wavelength of light.

 The above formulas assume that the slit width is very small compared to the wavelength of light, so that the

slits behave essentially like point sources of light.

𝝀
The angular spacing of the fringes (angular fringe width), θf, is then given by θf = [where θf <<1,]
𝒅

Fringes of equal thickness(FET):

The condition for maximum and minimum brightness of the fringes formed by light reflecting from a thin wedge-shaped
film are respectively
𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑚𝑎𝑥𝑖𝑚𝑎
𝟐µ𝒕𝒄𝒐𝒔 (𝒓 + 𝜽) = (𝟐𝒏 + 𝟏) 𝝀/𝟐 𝒏 = 𝟎, 𝟏, 𝟐, … . . . . . (𝟏)

𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 𝑚𝑖𝑛𝑖𝑚𝑎

𝟐µ𝒕𝒄𝒐𝒔 (𝒓 + 𝜽) = 𝒏𝝀 𝒏 = 𝟎, 𝟏, 𝟐, 𝟑, … … . . . (𝟐)
The above equations suggest that if the incident rays are parallel and monochromatic then µ, 𝑟, 𝜆 are constant. Under
this condition different order number of the fringes are controlled by the thickness t of the film. Hence a fringe of
particular order will lie on the locus of all points of the film having a constant thickness. This fringes are called fringes
of equal width or thickness.
Fringes of equal thickness are responsible for the rainbow colouring of thin films, such as soap bubbles, patches of oil
and gasoline on water, and films of oxides on metals; they are the cause of temper colour. Such fringes are often used
for determining the micro relief of thin plates and films and for making precise measurements with different kinds of
interferometers and other instruments. Newton’s rings are an example of fringes of equal thickness.

Define H-polaroid.

A film of polyvinyl alcohol (PVA) which is stretched to line up the complex molecules and then impregnated with
iodine. The film is then mounted between two thin glass plates named as H- polaroid.
How would you combine two linear vibrations to get circular vibration?
A circulating polarized light will produced by superposition of two mutually perpendicular linear vibrations of equal amplitudes
𝜋
but differing in phase by linear vibration at z=0.
2
𝜋
𝑥 = 𝑎 cos(𝜔𝑡 + ) 𝑎𝑛𝑑 𝑦 = 𝑎 cos 𝜔𝑡. Squaring and adding.
2
𝜋
So, circular vibration 𝑥 2 + 𝑦 2 = 𝑎2 𝑐𝑜𝑠 2 (𝜔𝑡 + ) + 𝑎2 𝑐𝑜𝑠 2 𝜔𝑡 = 𝑎2 𝑠𝑖𝑛2 𝜔𝑡 + 𝑎2 𝑐𝑜𝑠 2 𝜔𝑡 = 𝑎2
2

An EM waves 𝑬𝒙 (𝒛, 𝒕) = 𝑬𝑶𝒙 𝐜𝐨𝐬(𝒌𝒛 − 𝝎𝒕) 𝒂𝒏𝒅 𝑬𝒚 (𝒛, 𝒕) = 𝑬𝑶𝒚 𝐜𝐨𝐬(𝒌𝒛 − 𝝎𝒕 + 𝜹). Find the state of polarization in
𝝅 𝝅 𝝅
following cases (𝒊)𝜹 = 𝟎, 𝑬𝑶𝒙 ≠ 𝑬𝑶𝒚 ; (𝒊𝒊)𝜹 = , 𝑬𝑶𝒙 = 𝑬𝑶𝒚 ; (𝒊𝒊𝒊)𝜹 = − , 𝑬𝑶𝒙 = −𝑬𝑶𝒚 ; (𝒊𝒗)𝜹 = − , 𝑬𝑶𝒙 ≠ 𝑬𝑶𝒚 .
𝟐 𝟐 𝟐
(𝑖)𝛿 = 0, 𝐸𝑂𝑥 ≠ 𝐸𝑂𝑦 ⇒ 𝐸𝑥 = 𝐸𝑂𝑥 cos(𝑘𝑧 − 𝜔𝑡) 𝑎𝑛𝑑 𝐸𝑦 = 𝐸𝑂𝑦 cos(𝑘𝑧 − 𝜔𝑡)
𝐸𝑦 𝐸𝑂𝑦 𝐸 𝑂𝑦 𝐸𝑂𝑦
Dividing = ⇒ 𝐸𝑦 = 𝐸𝑥 . So, it is linearly polarized light in XY-plane, making an angle tan−1 with the x-axis.
𝐸𝑥 𝐸 𝑂𝑥 𝐸𝑂𝑥 𝐸 𝑂𝑥
𝜋 𝜋
(𝑖𝑖)𝛿 = , 𝐸𝑂𝑥 = 𝐸𝑂𝑦 ⇒ 𝐸𝑥 = 𝐸𝑂𝑥 cos(𝑘𝑧 − 𝜔𝑡) 𝑎𝑛𝑑 𝐸𝑦 = 𝐸𝑂𝑦 cos (𝑘𝑧 − 𝜔𝑡 + ) = −𝐸𝑂𝑥 sin(𝑘𝑧 − 𝜔𝑡)
2 2
Squaring and then adding, 𝐸𝑥 2 + 𝐸𝑦 2 = 𝐸𝑂𝑥 2 𝑐𝑜𝑠 2 (𝜔𝑡 − 𝑘𝑧) + 𝐸𝑂𝑥 2 𝑠𝑖𝑛 2 (𝜔𝑡 − 𝑘𝑧) = 𝐸𝑂𝑥 2
So, the light will be left circularly polarized light [left circular because [𝐸𝑥 = 𝐸𝑂𝑥 cos(𝑘𝑧 − 𝜔𝑡) & 𝐸𝑦 = −𝐸𝑂𝑥 sin(𝑘𝑧 − 𝜔𝑡)]
At 𝑧 = 0, 𝐸𝑥 = 𝐸𝑂𝑥 cos(−𝜔𝑡) = 𝐸𝑂𝑥 cos(𝜔𝑡) & 𝐸𝑦 = 𝐸𝑂𝑥 sin(𝜔𝑡) [∵ cos(−𝜃) = cos 𝜃 & sin(−𝜃) = − sin 𝜃]
At 𝑡 = 0 , 𝐸𝑥 = 𝐸𝑂𝑥 & 𝐸𝑦 = 0.
y
At Anticlockwise
𝑡 = ∆𝑡, 𝐸𝑥 = 𝐸𝑂𝑥 & 𝐸𝑦 = 𝐸𝑂𝑥 𝜔∆𝑡
[∵∆𝑡 is very small, cos 𝜔∆𝑡 ≈ 1 & sin 𝜔∆𝑡 ≈ 𝜔∆𝑡]
So, left circularly or anticlockwise.
x
𝑡=0
y

𝑡=0 x
Clockwise
𝜋
(𝑖𝑖𝑖)𝛿 = − , 𝐸𝑂𝑥 = −𝐸𝑂𝑦 ⇒ 𝐸𝑥 = 𝐸𝑂𝑥 cos(𝑘𝑧 − 𝜔𝑡) 𝑎𝑛𝑑 𝐸𝑦 = 𝐸𝑂𝑦 cos (𝑘𝑧 −
2
𝜋
𝜔𝑡 − ) = 𝐸𝑂𝑥 sin(𝑘𝑧 − 𝜔𝑡)
2
Squaring and then adding, 𝐸𝑥 2 + 𝐸𝑦 2 = 𝐸𝑂𝑥 2 𝑐𝑜𝑠 2 (𝑘𝑧 − 𝜔𝑡) + 𝐸𝑂𝑥 2 𝑠𝑖𝑛 2 (𝑘𝑧 − 𝜔𝑡) = 𝐸𝑂𝑥 2
So, the light will be left circularly polarized light.
At 𝑧 = 0, 𝐸𝑥 = 𝐸𝑂𝑥 cos(−𝜔𝑡) = 𝐸𝑂𝑥 cos(𝜔𝑡) & 𝐸𝑦 = 𝐸𝑂𝑥 sin(−𝜔𝑡) = −𝐸𝑂𝑥 sin(𝜔𝑡)
At 𝑡 = 0 , 𝐸𝑥 = 𝐸𝑂𝑥 & 𝐸𝑦 = 0.
At 𝑡 = ∆𝑡, 𝐸𝑥 = 𝐸𝑂𝑥 & 𝐸𝑦 = −𝐸𝑂𝑥 𝜔∆𝑡
So, right circularly or clockwise.

𝜋 𝜋
(𝑖𝑣)𝛿 = − , 𝐸𝑂𝑥 ≠ 𝐸𝑂𝑦 ⇒ 𝐸𝑥 = 𝐸𝑂𝑥 cos(𝑘𝑧 − 𝜔𝑡) 𝑎𝑛𝑑 𝐸𝑦 = 𝐸𝑂𝑦 cos (𝑘𝑧 − 𝜔𝑡 − ) = 𝐸𝑂𝑥 sin(𝑘𝑧 − 𝜔𝑡)
2 2
𝐸𝑥 𝐸𝑦
Now = cos(𝑘𝑧 − 𝜔𝑡)…….(i) and = sin(𝑘𝑧 − 𝜔𝑡)……..(ii)
𝐸 𝑂𝑥 𝐸𝑂𝑦
 2 2
𝐸𝑥 𝐸𝑦
Squaring (i) & (ii) and then adding, ( ) +( ) = 𝑐𝑜𝑠 2 (𝑘𝑧 − 𝜔𝑡) + 𝑠𝑖𝑛2 (𝑘𝑧 − 𝜔𝑡) = 1.
𝐸𝑂 𝑥 𝐸𝑂𝑦

So, the light will be right handed or clockwise elliptically polarized light.

A plane polarized light become circularly polarized light by passing through a crystal of thickness 𝟑 × 𝟏𝟎−𝟑 𝒄𝒎. Calculate
the difference in refractive indices for E ray and O ray if 𝝀 = 𝟔𝟎𝟎𝟎Å.
𝜆
A plane polarized light become circularly polarized if it is pass through a (quarter wave) plate. Now path difference between O
4
𝜆
& E ray= 𝑑(𝑛𝑜 − 𝑛𝑒 ) . ∴ 𝑑(𝑛𝑜 − 𝑛𝑒 ) =
4
[𝑑=thickness=3 × 10−3 𝑐𝑚, 𝑛𝑜 , 𝑛𝑒 refractive indices for O & E ray, 𝜆 = 6000Å = 6000 × 10−8 cm]
6000×10−8
(𝑛𝑜 − 𝑛𝑒 ) = = 5 × 10−3 = 0.005.
4×3×10−3

How population inversion is obtained in practical lasers?

Population inversion cannot be achieved by a two-level system. You need either three or four-level systems. What you must do is
pump the atoms from the lowest energy to the highest energy, not necessarily using an optical transition, but even via an electrical
discharge, perhaps. Also, you must ensure that the transition from the high energy to the middle energy level (meta stable state) is
fast, and radiation less. You can also make a laser using a four-level system, in which case two of the transitions between levels
must be fast and radiation less, with one of the others being the lasing transition.

Discuss what are the basic difference between spontaneous and stimulated emission.