Question

A double slit experiment is performed with light of wavelength $500nm$. A thin film of thickness $2\mu m$ and refractive index $1.5$ is introduced in the path of the upper beam. The location of the central maximum will

• A. remain unshifted
• B. shift downward by nearly two fringes
• C. shift upward by nearly two fringes
• D. shift downward by its fringes

Answer 1

Answer: (C)

In Young's Double Slit Experiment (YDSE) be the position of central bright fringe on the screen in absence of any film, because the $S_{1}O=S_{2}O$.

When we introduce a film of thickness $t$ and refractive index $\mu$, an additional path difference equal to $(\mu t-t)=(\mu -1)t$ is introduced. The optical path of upper beam becomes longer. For path difference on screen to be zero, path from lower slit $S_2$ should also be more. Thus, central bright fringe wire be located some what at $O^{\prime }$ as $S_2{O}^{\prime }>S_{2}O$.
$\therefore$ The fringe pattern shifts upwards.
Now as a change in path difference of $\lambda$ corresponds to a change in position on the screen by $\beta =\frac{D}{d}\lambda$
Change in optical path difference $\Delta y=(\mu -1)t\times \frac{D}{d}$
$\Delta y=(1.5-1)\left(2\times 10^{-6}\right)\frac{D}{d}$
$\Delta y=\frac{1}{2}\times 2\times 10^{-6}\frac{D}{d}=\frac{D}{d}\times 10^{-6}m$
Fringe width $=\frac{D}{d}\lambda =\frac{D}{d}\times 500\times 10^{-9}m=\beta$
Clearly, $\frac{\Delta y}{\beta }=\frac{\frac{D}{d}\times 10^{-6}}{\frac{D}{d}\times 500\times 10^{-9}}=2$
$=\frac{10^{-6}}{500\times 10^{-9}}=\frac{10^{-6}\times 10^9}{500}=\frac{10^3}{500}$
$\therefore \Delta y=2\beta$