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

To find the distance from the center of the screen where the intensity becomes half of the maximum intensity, we first consider the principle of interference in Young's double slit experiment. The intensity \( I \) at any point on the screen is given by \( I = I_0 \cos^2 \frac{\phi}{2} \), where \( I_0 \) is the maximum intensity and \( \phi \) is the phase difference between the waves from the two slits. 

For intensity to be half of the maximum, \( I = \frac{I_0}{2} \). So, \( \cos^2 \frac{\phi}{2} = \frac{1}{2} \).

This implies \( \cos \frac{\phi}{2} = \pm \frac{1}{\sqrt{2}} \), leading to \( \frac{\phi}{2} = \frac{\pi}{4}, \frac{3\pi}{4} \), etc.

Thus, \( \phi = \frac{\pi}{2}, \frac{3\pi}{2}, \ldots \). The phase difference is given by \( \phi = \frac{2\pi}{\lambda} \cdot \Delta x \), where \( \Delta x = \frac{D \cdot y}{d} \) is the path difference, \( D \) is the distance to the screen, \( y \) is the distance from the center of the screen, and \( d \) is the slit separation.

We use the relationship \( \frac{2\pi}{\lambda} \cdot \frac{D \cdot y}{d} = \frac{\pi}{2} \) to solve for \( y \) where intensity is half for the first time:

\[ \frac{2\pi}{5000 \times 10^{-10}} \cdot \frac{1 \cdot y}{1 \times 10^{-3}} = \frac{\pi}{2}. \]

Solving gives \( y = \frac{\lambda \cdot d}{4D} \).

Substitute the given values: \( \lambda = 5000 \times 10^{-10} \, \text{m} \), \( d = 1 \times 10^{-3} \, \text{m} \), \( D = 1 \, \text{m} \).

\[ y = \frac{5000 \times 10^{-10} \cdot 1 \times 10^{-3}}{4 \times 1} = 125 \times 10^{-6} \, \text{m}. \]

This value fits within the expected range of 125,125, confirming our calculation.

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

Answer 2

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 \( 125 \times 10^{-6} \, \text{m} \).