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) 1.184 μm

Answer 2

(a) The fringe width in a medium is given by $\beta_l = \frac{\lambda_l D}{d}$, where $\lambda_l = \frac{\lambda_{air}}{n_l}$. Substituting the values, $\beta_l = \frac{(\frac{6300 \times 10^{-10}}{1.33}) \times 1.33}{1 \times 10^{-3}} = 0.63 \times 10^{-3}$ m = 0.63 mm. (b) The optical path difference introduced by the glass sheet is $\Delta P = (n_g - n_l)t$. For an adjacent minimum to be on the axis, the central maximum must shift by half a fringe width, so $\Delta y = \frac{1}{2} \beta_l$. The shift is also given by $\Delta y = \frac{D}{d} \Delta P$. Equating these, $\frac{D}{d} (n_g - n_l)t = \frac{1}{2} \frac{\lambda_l D}{d}$. This simplifies to $(n_g - n_l)t = \frac{1}{2} \lambda_l = \frac{1}{2} \frac{\lambda_{air}}{n_l}$. Solving for $t = \frac{\lambda_{air}}{2 n_l (n_g - n_l)} = \frac{6300 \times 10^{-10}}{2 \times 1.33 \times (1.53 - 1.33)} \approx 1.184 \times 10^{-6}$ m = 1.184 μm.