In a Young's double slit experiment set up, the two slits are kept 0.4. mm apart and screen is placed at 1 m from slits. If a thin transparent sheet of thickness $$20\mu m$$ is introduced in front of one of the slits then center bright fringe shifts by 20 mm on the screen. The refractive index of transparent sheet is given by $$\frac{\alpha}{10}$$, where $$\alpha$$ is __________.

Slit separation $$d = 0.4$$ mm = $$0.4 \times 10^{-3}$$ m, screen distance $$D = 1$$ m, sheet thickness $$t = 20 \mu m = 20 \times 10^{-6}$$ m, fringe shift = 20 mm = $$20 \times 10^{-3}$$ m.

Fringe shift due to thin film: $$\Delta = \frac{(\mu - 1)tD}{d}$$

$$20 \times 10^{-3} = \frac{(\mu - 1) \times 20 \times 10^{-6} \times 1}{0.4 \times 10^{-3}}$$

$$20 \times 10^{-3} = \frac{(\mu-1) \times 20 \times 10^{-6}}{4 \times 10^{-4}} = (\mu - 1) \times 50 \times 10^{-3}$$

$$\mu - 1 = \frac{20}{50} = 0.4$$

$$\mu = 1.4 = \frac{14}{10}$$

So $$\alpha = 14$$.

The answer is 14.