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 determine the distance from the screen's center at which the intensity reduces to half of its maximum value, we apply the principle of interference in Young's double-slit experiment. The intensity \( I \) at any point on the screen is expressed by \( I = I_0 \cos^2 \frac{\phi}{2} \), where \( I_0 \) represents the maximum intensity and \( \phi \) denotes the phase difference between the waves originating from the two slits.

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

This condition implies \( \cos \frac{\phi}{2} = \pm \frac{1}{\sqrt{2}} \), which leads to phase differences \( \frac{\phi}{2} = \frac{\pi}{4}, \frac{3\pi}{4} \), and so forth.

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

We utilize the relationship \( \frac{2\pi}{\lambda} \cdot \frac{D \cdot y}{d} = \frac{\pi}{2} \) to calculate \( y \), the distance from the center where the intensity first becomes half of the maximum:

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

Solving this equation yields \( y = \frac{\lambda \cdot d}{4D} \).

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

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

This calculated value of \( 125 \times 10^{-6} \, \text{m} \) is consistent with expected results, validating the computation.

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