Question

In Young's double slit experiment, monochromatic light of wavelength 5000 \, \text{Å} is used. The slits are 1.0 mm apart and the screen is placed 1.0 m away from the slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is \[\_ \times 10^{-6} \, \text{m}.\]

Answer 1

Correct Answer: 125

Given: - Wavelength of light: \( \lambda = 5000 \, \text{\AA} = 5000 \times 10^{-10} \, \text{m} \) - Distance between slits: \( d = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m} \) - Distance between the slits and the screen: \( D = 1.0 \, \text{m} \)

Step 1: Calculating the Fringe Width

The fringe width \( \beta \) in Young’s double-slit experiment is given by:

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

Substituting the given values:

\[ \beta = \frac{5000 \times 10^{-10} \times 1.0}{1.0 \times 10^{-3}} \, \text{m} \] \[ \beta = 5 \times 10^{-3} \, \text{m} = 5 \, \text{mm} \]

Step 2: Finding the Distance Where Intensity is Half of Maximum

The intensity becomes half of the maximum intensity at the position of the first secondary maximum. The position \( y \) where this occurs is given by:

\[ y = \frac{\beta}{4} \]

Substituting the value of \( \beta \):

\[ y = \frac{5 \times 10^{-3}}{4} \, \text{m} \] \[ y = 1.25 \times 10^{-3} \, \text{m} = 125 \times 10^{-6} \, \text{m} \]

Conclusion:

The distance from the centre of the screen where the intensity becomes half of the maximum intensity for the first time is \( 125 \times 10^{-6} \, \text{m} \).