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 $(\lambda=4000 \AA)$ 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
