Young Double Slits Experiment Derivation: Overview, Questions, Preparation

Two coherent light sources positioned at a small distance apart are used in Young's Double Slit Experiment, typically just a few orders of magnitude greater than the wavelength of light.

Derivation:

The distance between the slit and the screen is D, and the width of the slit is d.

The light waves from s₁ and s₂ must travel separate distances to meet P while the split separation (d) and the frame distance (D) are kept unchanged. It means that there is a direct gap between the two light waves from s₁ and s₂ in Young's double-slit experiment.

Fringes with maximum intensity

For n = 0, the light fringe is known as the central fringe. The fringes of the higher-order are symmetrically positioned along the central fringe. The position of the bright fringe is given by the nth position.

Y (brightness) = (nλ\d)D (n = 0, ±1, ±2, . . . .)

Fringes with minimum intensity

Y (dark) = (2n−1)λD/2d (n = ±1,±2,…..)

It is the distance between two adjacent bright or adjacent dark fringes.

Let us assume we are considering two bright fringes at positions n and n+1, respectively.

So fringe width = ((n+1)λ\d)D - (nλ\d)D = (λ\d)D

The intensity of fringes in Young’s Double Slit Experiment

For two coherent sources, the resultant intensity at point P is given by

I = I₁ + I₂ + 2 √(I₁ . I₂) cos φ

I₁=I₂ = I

Putting value, we get: 

φ= 4. I . cos²(φ/2)

For maximum intensity = φ= 2npie

For minimum intensity = φ = 2(n-1)pie

Young’s experiment in Class 12

Young’s experiment is part of wave optics, so in chapter wave optics, you will get to learn the derivation in detail. The weightage of this chapter is 6-7 marks.

Example 1: In Young's double-slit experiment, if the slit widths are in the ratio 1:9, then the balance of the intensity at minima to that at maxima will be? 

Solution: Slit width ratio = 1 : 9 Since slit width ratio is the ratio of intensity and intensity∝(amplitude)2 

I₁ : I₂ = 1 : 9 

a₁²:a₂²= 1:9

a₁:a₂ = 1:3

I_max= (a₁+a₂)²

I_min=(a₁-a₂)²

I_max:I_min= 1:4

Example 2: Two coherent light sources S₁ and S₂(I=6000 A) are 1mm apart from each other. The screen is placed at a 25 CM from the sources, the width of the fringes on the screen is?

Solution:

ƀ=6000 * 10^-10 *25*10^-2/ 10^-3

= 150*10^-6   

Example 3: The width of the fringes obtained from the light of wavelength 500 nm is 3.6 mm. What is the fringe width if the apparatus is immersed in a liquid of refractive index 1.2?

Solution:

β’ = λ’D/d

β’/ β = λ’/λ

β’ = β/μ

= 3.6 mm/1.2

Image Courtesy: NCERT

Q: What is the intention of Young's experiment with double slits?

Q: What gives Young's double-slit experiment a dark spot on the screen?

Q: What was proved by Young's experiment?

Q: When you raise the distance between the slits, what happens?

Q: What hypothesis is most successfully demonstrated by Young's double-slit experiment?