In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness t and refractive index n(=1.5), the central fringe shifts by 0.2 cm. The value of t is____ cm

Let
$$d = 0.1 \text{ cm} = 1.0 \times 10^{-3} \text{ m}$$ be the slit separation,

$$D = 50 \text{ cm} = 0.5 \text{ m}$$ be the distance of the screen from the slits,

shift of the central fringe $$y_0 = 0.2 \text{ cm} = 0.2 \times 10^{-2} \text{ m} = 2.0 \times 10^{-3} \text{ m},$$

and refractive index of the sheet $$n = 1.5 \; \Rightarrow \; (n-1) = 0.5.$$

Insertion of a transparent sheet of thickness $$t$$ in front of one slit introduces an additional optical path difference
$$\Delta = (n-1)t.$$

The displacement of the central fringe on the screen is given by the formula
$$y_0 = \frac{(n-1)\,t\,D}{d} \quad -(1).$$

Rearranging $$(1)$$ for $$t$$:
$$t = \frac{y_0\,d}{(n-1)\,D} \quad -(2).$$

Substituting the numerical values into $$(2)$$:
$$t = \frac{\bigl(2.0 \times 10^{-3}\bigr)\bigl(1.0 \times 10^{-3}\bigr)}{0.5 \times 0.5}$$

$$t = \frac{2.0 \times 10^{-6}}{0.25}$$

$$t = 8.0 \times 10^{-6} \text{ m}.$$

Converting to centimetres:
$$t = 8.0 \times 10^{-6} \text{ m} \times 100 \frac{\text{cm}}{\text{m}} = 8.0 \times 10^{-4} \text{ cm}.$$

Therefore, the required thickness is $$8 \times 10^{-4} \text{ cm},$$ which corresponds to Option C.