Question

In Young’s double slit experiment, two slits are placed 2 mm from each other. The interference pattern is observed on a screen placed 2 m from the plane of the slits. Then the fringe width for a light of wavelength 400 nm is:

• A. \( 0.4 \times 10^{-6} \text{ m} \)
• B. \( 4 \times 10^{-6} \text{ m} \)
• C. \( 0.4 \times 10^{-3} \text{ m} \)
• D. \( 400 \text{ m} \)

Hint:

In Young’s double slit experiment, the fringe width \( \beta = \frac{\lambda D}{d} \) directly depends on the wavelength \( \lambda \) and screen distance \( D \), and inversely depends on the slit separation \( d \). Always ensure correct unit conversions before substituting values.

Answer 1

The Correct Option is C

Step 1: Understanding the Fringe Width Formula In Young’s double slit experiment, the fringe width \( \beta \) is given by: \[ \beta = \frac{\lambda D}{d} \] where: - \( \lambda = 400 \) nm = \( 400 \times 10^{-9} \) m (wavelength of light), - \( D = 2 \) m (distance between slits and screen), - \( d = 2 \) mm = \( 2 \times 10^{-3} \) m (distance between slits). 
Step 2: Calculating Fringe Width Substituting the given values: \[ \beta = \frac{(400 \times 10^{-9}) \times 2}{2 \times 10^{-3}} \] \[ \beta = \frac{800 \times 10^{-9}}{2 \times 10^{-3}} \] \[ \beta = 0.4 \times 10^{-3} \text{ m} \] 
Step 3: Conclusion Thus, the fringe width is \( 0.4 \times 10^{-3} \) m.