Question

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

• A. Remain unstated
• B. Shift downward by neary two fringes
• C. Shift upward by nearly two fringes
• D. Shift downward by its fringes

Answer 1

Answer: (C)

Shift upward by nearly two fringes

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 p, an additional path difference equal to $(\mu t-t)=(\mu -1)t$ is introduced. The optical path of upper bean 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 $S_2{O}^{\prime }>S_{2}O$.
$\therefore$ 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 $\Delta y=(\mu -1)t\times \frac{D}{d}$
$\Delta y=(1.5-1)(2\times {10}^{-6})\frac{D}{d}$
$\Delta y=\frac{1}{2}\times 2\times {10}^{-6}\frac{D}{d}=\frac{D}{d}\times {10}^{-6}m$
Fringe width $=\frac{D}{d}\lambda =\frac{D}{d}\times 100\times {10}^{-9}m=\beta$
Clearly, $\frac{\Delta y}{\beta }=\frac{\frac{D}{d}\times {10}^{-6}}{\frac{D}{d}\times 500\times {10}^{-9}}=2$
$=\frac{{10}^{-6}}{500\times {10}^{-9}}=\frac{{10}^{-6}\times {10}^9}{500}=\frac{{10}^3}{500}$
$\therefore \Delta y=2\beta$