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

Young’s Double Slit Experiment – Fringe Width Calculation 

The fringe width \( w \) in Young’s double slit experiment is given by the formula:

\[ w = \frac{\lambda D}{d} \]

Where:

• \( \lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \) (wavelength of light)
• \( D = 2 \, \text{m} \) (distance between slits and screen)
• \( d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \) (distance between the slits)

Substitute the values:

\[ w = \frac{400 \times 10^{-9} \times 2}{2 \times 10^{-3}} = \frac{800 \times 10^{-9}}{2 \times 10^{-3}} \]

Simplify:

\[ w = 0.4 \times 10^{-3} \, \text{m} = \boxed{0.4 \, \text{mm}} \]

Conclusion:

The fringe width in the given setup is \( \boxed{0.4 \, \text{mm}} \).

Answer 2

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.