Question

In a Young’s double slit experiment, a combination of two glass wedges $ A $ and $ B $, having refractive indices 1.7 and 1.5, respectively, are placed in front of the slits, as shown in the figure. The separation between the slits is $ d = 2 \text{ mm} $ and the shortest distance between the slits and the screen is $ D = 2 \text{ m} $. Thickness of the combination of the wedges is $ t = 12 \, \mu\text{m} $. The value of $ l $ as shown in the figure is 1 mm. Neglect any refraction effect at the slanted interface of the wedges. Due to the combination of the wedges, the central maximum shifts (in mm) with respect to 0 by ____

Hint:

The central fringe shift in a double slit experiment with wedges depends on the difference in refractive indices, thickness of wedges, slit separation, and screen distance.

Answer 1

Step 1: The shift of the central maximum due to the wedges is given by the phase difference introduced between the two paths due to different refractive indices.
Step 2: The path difference introduced by the wedges is \[ \Delta = (n_A - n_B) \times t = (1.7 - 1.5) \times 12 \times 10^{-6} \text{ m} = 0.2 \times 12 \times 10^{-6} = 2.4 \times 10^{-6} \text{ m} \]
Step 3: The shift in the central maximum on the screen is given by \[ \Delta x = \frac{D \Delta}{d} \] where \( D = 2 \text{ m} \), \( \Delta = 2.4 \times 10^{-6} \text{ m} \), \( d = 2 \text{ mm} = 2 \times 10^{-3} \text{ m} \).
Step 4: Calculate the shift: \[ \Delta x = \frac{2 \times 2.4 \times 10^{-6}}{2 \times 10^{-3}} = \frac{4.8 \times 10^{-6}}{2 \times 10^{-3}} = 2.4 \times 10^{-3} \text{ m} = 0.0024 \text{ m} = 2.4 \text{ mm} \] Step 5: However, the shift must be calculated using the wedge length \( l = 1 \text{ mm} \), the effective phase change is proportional to \( l \), so the actual shift: \[ \Delta x = \frac{D}{d} (n_A - n_B) \times t \times \frac{l}{t} = \frac{D}{d} (n_A - n_B) \times l = \frac{2}{2 \times 10^{-3}} \times 0.2 \times 10^{-3} = 1000 \times 0.2 \times 10^{-3} = 0.2 \text{ mm} \] Thus, the central maximum shifts by \(0.2 \text{ mm}\) on the screen.