Question

In Youngs double slit experiment, the central fringe of interference pattern produced by light of wavelength $ 6000\,\overset{\text{o}}{\mathop{\text{A}}}\, $ shifts to the position of the 5th bright fringe on introducing a thin glass plate of refractive index of 1.50 in front of one of the slits. The thickness of the plate would be:

• A. $ \text{6 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{cm} $
• B. $ \text{4 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{cm} $
• C. $ \text{5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{cm} $
• D. $ \text{6 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-6}}}\text{cm} $

Answer 1

The Correct Option is A

The shift in the fringe pattern – 4λD/d

Here, λ = 6000x10^-10m

Now, shift due to glass plate= (μ-1)tD/d

Now, equating both the equations–

5λD/d =  (μ-1)tD/d

So, t=  5λ/(μ-1)

t= 5x6000x10^-10/ (1.5-1) = 6x10^-6m = 6x10^-4cm
The light fringe is referred to as the core fringe for N=0. Along the core fringe, the higher-order fringes are asymmetrically placed. The nth fringe indicates where the brilliant fringe is located.

Y(brightness) = (nπ\d)D (n=0, n=+1 -1, +2 -2….)

Fringe’s with minimum intensity: Y(dark) = (2n-1)πD/2d (n=0, n=+1 -1, +2 -2….)

Fringe width is the distance between two adjacent bright and adjacent dark fringes.

Let's consider the two fringes between position n and n+1

So, fringe width = ((n+1 π\d ) D - (nπ\d)D = πD/ d

The distance of the nth bright fringe from the center is

x_n = nλD/d

The distance of the nth dark fringe from the center is:

x_n = (2n+1)λD/2d