### Wave Optics Question 6

#### Question 6 - 30 January - Shift 1

In Young’s double slit experiment, two slits $S_1$ and $S_2$ are ’d’ distance apart and the separation from slits to screen is D (as shown in figure). Now if two transparent slabs of equal thickness $0.1 mm$ but refractive index 1.51 and 1.55 are introduced in the path of beam $(\lambda=4000 \mathring{A})$ from $S_1$ and $S_2$ respectively. The central bright fringe spot will shift by number of fringes.

## Show Answer

#### Answer: (10)

#### Solution:

#### Formula: YDSE

Path difference at $P$ be $\Delta x$

$ \begin{aligned} \Delta x & =(\mu_2-\mu_1) t \\ & =(1.55-1.51) 0.1 mm \\ & =0.04 \times 10^{-4} \end{aligned} $

$\Delta x=4 \times 10^{-6}=4 \mu m$

$y=\frac{\Delta xD}{d}=4 \times 10^{-6} \frac{D}{d}$

y is the distance of central maxima from geometric center

$\underset{(\beta)}{\text{ fringe width }}=\frac{\lambda D}{d}=4 \times 10^{-6} m \frac{D}{d}=4 \mu m \frac{D}{d}$

$\therefore \quad$ Central bright fringe spot will shift by ’ $x$ '

Number of shift $=\frac{y}{\beta}$

$=\frac{4 \times 10^{-6} D / d}{4 \times 10^{-7} D / d}=10$ Ans