Question

If a mica sheet of thickness $t$ and refractive index $\mu$ is placed in the path of one of interfering beams in a double slit experiment, then displacement of fringes will be :

• A. $\frac{D}{d}\mu t$
• B. $\frac{D}{d}(\mu -1)t$
• C. $\frac{D}{d}(\mu +1)t$
• D. $\frac{D}{d}({\mu }^2-1)t$

Answer 1

Answer: (B)

We can realise the situation as shown. Geometric path difference between $S_{2}P$ and $S_{1}P$ is
$\Delta x_1=S_{2}P-S_{1}P=\frac{yd}{D}$

where $d$ is distance between the slits and $D$ the distance between source and screen.
Path difference produced by the mica sheet
$\Delta x_2=(\mu -1)t$
Therefore, net path difference between the two rays is
$\Delta x=\Delta x_1-\Delta x_2$
or $\Delta x=\frac{yd}{D}-(\mu -1)t$
For $n$th maxima on upper side,
$\Delta x=n\lambda$
$\therefore \frac{yd}{D}-(\mu -1)t=n\lambda$
$\therefore y=\frac{n\lambda D}{d}+\frac{(\mu -1)tD}{d}$
Earlier it was $\frac{n\lambda D}{d}$.
$\therefore$ Displacement of fringes $=\frac{(\mu -1)tD}{d}$