Question

In Young's experiment, monochromatic light
through a single slit S is used to illuminate the two
slits S1 and S2. Interference fringes are obtained on
a screen. The fringe width is found to be w. Now a
thin sheet of mica (thickness t and refractive index
m) is placed near and in front of one of the two slits.
Now the fringe width is found to be w¢, then :
(1) w¢ = w/m
(2) w¢ = wm
(3) w¢ =(m 1) tw
(4) w¢ = w

Answer 1

Given that,

fringe width =  w

Thickness = t

Refractive index = μ

We know that,

The fringe of width is

$w=\dfrac{D\lambda}{d}$...(I)

The distance of fringes are

$y_{n}=\dfrac{D}{d}(n\lambda+(\mu-1)t)$

When thin sheet of mica is placed near and in front of one of the two slits.

We need to calculate the width of fringe

Using formula of width of fringe

$w'=y_{n+1}-y_{n}$

Put the value into the formula

$w'=\dfrac{D}{d}((n+1)\lambda+(\mu-1)t)-\dfrac{D}{d}(n\lambda+(\mu-1)t)$

$w'=\dfrac{D\lambda}{d}$

From equation (I)

$w'=w$

Hence, The fringe width always same.

(4) is correct option.