Question

A double slit experiment is performed with light of wavelength 500 nm. 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 10 fringes

Answer 1

The correct option is C Shift upward by nearly two fringes

$Shift=\frac{D}{d}(\mu -1)t=\frac{D}{d}[1.5-1]\times 2\times {10}^{-6}=\frac{D}{d}\times {10}^{-6}$
Now, fringe width $\omega =\frac{D}{d}\times 500\times {10}^{-9}$
$\therefore \frac{shift}{\omega }=\frac{\frac{D}{d}\times {10}^{-6}}{\frac{D}{d}\times 5\times {10}^{-7}}=2$
$\therefore$ $shift=2\omega$, since the film is kept in the path of upper beam, the shift will be upward.