Question

A transparent slab of thickness t and refractive index $\mu$is inserted in front of upper slit of YDSE apparatus. The wavelength of light used is $\lambda$. Assume that there is no absorption of light by the slab. Select the correct statement (s) :

• A. The intensity of dark fringes will be 0, if slits are identical
• B. The change in optical path due to insertion of plate is $\mu t$
• C. The change in optical path due to insertion of plate is \(\mu - 1)t\)
• D. For producing intensity zero at the centre of screen, the thickness could be $\frac{5\lambda }{2(\mu -1)}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A The intensity of dark fringes will be 0, if slits are identical
C The change in optical path due to insertion of plate is \(\mu - 1)t\)
D For producing intensity zero at the centre of screen, the thickness could be $\frac{5\lambda }{2(\mu -1)}$

$\Delta x$ at P:
$\Delta x=(S_{1}P-t{)}_{air}+t_{medium}-S_2{P}_{air}$

$=[S_{1}P-S_{2}P+\mu t-t{]}_{air}$
$=(\mu -1)t$
Earlier, $\Delta x$ at $P=S_{1}P-S_{2}P=0$
So, change in optical path due to insertion of slab
$=(\mu -1)t$
For intensity to be zero at P, we have
$\Delta x=\frac{(2n-1)\lambda }{2}\Rightarrow (\mu -1)t=\frac{(2n-1)\lambda }{2}forn=3,t=\frac{5\lambda }{2(\mu -1)}$