Question

A double slit apparatus is immersed in a liquid of refractive index 1.33. It has slit separation of 1 mm and distance between the plane of slits and the screen is 1.33 m. The slits are illuminated by a parallel beam of light whose wavelength in air is 6300 Å. (a) Calculate the fringe width. (b) One of the slits of the apparatus is covered by a thin glass sheet of refractive index 1.53. Find the smallest thickness of the sheet to bring the adjacent minima on the axis.

Answer 1

(a) 0.63 mm, (b) 1.575 µm

Answer 2

(a) The fringe width in a medium is given by $\beta = \frac{\lambda_0 D}{n d}$. Given $\lambda_0 = 6300$ Å $= 6.3 \times 10^{-7}$ m, $n = 1.33$, $D = 1.33$ m, $d = 1$ mm $= 10^{-3}$ m. $\beta = \frac{(6.3 \times 10^{-7} \text{ m}) \times (1.33 \text{ m})}{1.33 \times (1 \times 10^{-3} \text{ m})} = 6.3 \times 10^{-4} \text{ m} = 0.63 \text{ mm}$.

(b) For the axis to become a minimum, the central maximum must shift by half a fringe width ($\beta/2$). The shift of the central maximum is given by $\Delta y = -\frac{(n_g - n_m)t D}{n_m d}$. The fringe width in the medium is $\beta = \frac{\lambda_0 D}{n_m d}$. Equating the shift to half the fringe width: $\left|-\frac{(n_g - n_m)t D}{n_m d}\right| = \frac{1}{2} \frac{\lambda_0 D}{n_m d}$. This simplifies to $(n_g - n_m)t = \frac{1}{2} \lambda_0$, so $t = \frac{\lambda_0}{2(n_g - n_m)}$. Given $\lambda_0 = 6.3 \times 10^{-7}$ m, $n_g = 1.53$, $n_m = 1.33$. $t = \frac{6.3 \times 10^{-7} \text{ m}}{2 \times (1.53 - 1.33)} = \frac{6.3 \times 10^{-7}}{2 \times 0.20} = 1.575 \times 10^{-6} \text{ m} = 1.575 \text{ µm}$.