Question

A double slit experiment is performed with light of wavelength $500 \,nm$. A thin film of thickness $2\, μm$ 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 $O'$ as $S_{2} O'>S_{2} O$.
$\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, $\frac{\Delta y}{\beta} =\frac{\frac{D}{d} \times 10^{-6}}{\frac{D}{d} \times 500 \times 10^{-9}}=2$
$=\frac{10^{-6}}{500 \times 10^{-9}}=\frac{10^{-6} \times 10^{9}}{500}=\frac{10^{3}}{500}$
$\therefore \Delta y =2 \beta$