Question

Young's double slit experiment is conducted in a medium of refractive index 4/3. Light of 600 nm wavelength falls on the slits having 0.45 mm separation. The lower slit $S_2$ is covered by a thin glass sheet of thickness 10.4 $\mu$m and refractive index 1.5. The interference pattern is observed a screen placed 1.5 m from the slits as shown in the figure. Find the location of central maxima (bright fringe with zero path difference on the y-axis) in mm [All wavelength in the problem are for the given medium of refractive index 4/3. Ignore dispersion.]

Answer 1

-4.33

Answer 2

The central maximum occurs at the point where the optical path difference between the waves from the two slits is zero. Let $n_m$ be the refractive index of the medium, $n_g$ be the refractive index of the glass sheet, $t$ be the thickness of the glass sheet, $d$ be the separation between the slits, and $D$ be the distance between the slits and the screen. Let $y$ be the vertical position of a point P on the screen from the central point O (directly opposite the midpoint of the slits).

The optical path difference between the waves from $S_2$ and $S_1$ at point P is given by: $\Delta OP = n_m (r_2 - r_1) + (n_g - n_m)t$ where $r_1$ and $r_2$ are the geometric path lengths from $S_1$ and $S_2$ to P, respectively.

For a point P at a vertical distance $y$ from O, the geometric path difference $r_2 - r_1$ is approximately $yd/D$ for small angles, assuming $S_1$ is the upper slit and $S_2$ is the lower slit, and y is measured upwards from O. From the figure, $S_1$ is above $S_2$.

So, $\Delta OP = n_m \frac{yd}{D} + (n_g - n_m)t$.

For the central maximum, $\Delta OP = 0$: $n_m \frac{yd}{D} + (n_g - n_m)t = 0$ $n_m \frac{yd}{D} = -(n_g - n_m)t = (n_m - n_g)t$ $y = \frac{(n_m - n_g)t D}{n_m d}$.

Given values: $n_m = 4/3$ $n_g = 1.5 = 3/2$ $t = 10.4 \, \mu\text{m} = 10.4 \times 10^{-6} \, \text{m}$ $d = 0.45 \, \text{mm} = 0.45 \times 10^{-3} \, \text{m}$ $D = 1.5 \, \text{m}$

Calculate $n_m - n_g$: $n_m - n_g = \frac{4}{3} - \frac{3}{2} = \frac{8 - 9}{6} = -\frac{1}{6}$.

Substitute the values into the formula for $y$: $y = \frac{(-1/6) \times (10.4 \times 10^{-6} \, \text{m}) \times (1.5 \, \text{m})}{(4/3) \times (0.45 \times 10^{-3} \, \text{m})}$ $y = \frac{-\frac{1}{6} \times 10.4 \times 1.5}{\frac{4}{3} \times 0.45} \times \frac{10^{-6}}{10^{-3}}$ $y = \frac{-\frac{15.6}{6}}{\frac{1.8}{3}} \times 10^{-3}$ $y = \frac{-2.6}{0.6} \times 10^{-3}$ $y = -\frac{26}{6} \times 10^{-3} = -\frac{13}{3} \times 10^{-3}$ $y \approx -4.333... \times 10^{-3} \, \text{m}$.

Convert to mm: $y \approx -4.333... \, \text{mm}$.

The negative sign indicates that the central maximum is located below the point O.