Question

In a biprism experiment, the slit separation is 1 mm. Using monochromatic light of wavelength 5000Å, an interference pattern is obtained on the screen. Where should the screen be moved, so that the change in fringe width is \( 12.5 \times 10^{-5} \, \text{m} \)?

• A. Away or towards the slit by 25 cm
• B. Away or towards the slit by 12.5 cm
• C. Away from the slit by 5 cm
• D. Towards the slit by 10 cm

Hint:

In interference patterns, the fringe width is directly proportional to the distance between the slits and the screen, and inversely proportional to the slit separation.

Answer 1

The Correct Option is A

Step 1: Using the fringe width formula.
The fringe width \( \beta \) is given by: \[ \beta = \frac{\lambda D}{d} \] Where: - \( \lambda = 5000 \, \text{Å} = 5 \times 10^{-7} \, \text{m} \) (wavelength), - \( D = 1 \, \text{m} \) (distance between the slits and screen), - \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \) (slit separation). Step 2: Calculating the fringe shift.
The fringe width is \( \beta = \frac{(5 \times 10^{-7}) \times 1}{1 \times 10^{-3}} = 5 \times 10^{-4} \, \text{m} \). Now, if the fringe width changes by \( 12.5 \times 10^{-5} \, \text{m} \), the screen must be moved by: \[ \text{Change in distance} = \frac{12.5 \times 10^{-5}}{5 \times 10^{-4}} \times 1 \, \text{m} = 0.25 \, \text{m} = 25 \, \text{cm} \] Thus, the screen should be moved by 25 cm, corresponding to option (A).