Question

In the Young's double slit experiment, the intensity of light at a point on the screen where the path difference $\lambda$ is K,( $\lambda$ being the wavelength of light used). The intensity at a point where the path difference is ${\lambda}/4 $ will be

• A. K
• B. K/4
• C. K/2
• D. zero

Answer 1

The Correct Option is C

In the Young's double slit experiment, the intensity \( I \) at a point on the screen depends on the path difference \( \Delta x \) between the two waves. The intensity can be expressed as:

I = I_0 \cos^2\left(\frac{\pi \Delta x}{\lambda}\right)

where \( I_0 \) is the maximum intensity, \( \lambda \) is the wavelength of the light used, and \( \Delta x \) is the path difference.

Given:

• At path difference \( \Delta x = \lambda \), intensity is \( K \).
• We need to find the intensity when \( \Delta x = \frac{\lambda}{4} \).

Step-by-step Solution:

1. Case 1: \(\Delta x = \lambda\)
   • The intensity \( I_1 \) is: I_1 = I_0 \cos^2\left(\frac{\pi \lambda}{\lambda}\right) = I_0 \cos^2(\pi) = I_0 \times 0 = 0 .
2. Case 2: \(\Delta x = \frac{\lambda}{4}\)
   • The intensity at this path difference, \( I_2 \), is given by: I_2 = I_0 \cos^2\left(\frac{\pi}{4}\right) = I_0 \left(\frac{1}{\sqrt{2}}\right)^2 = I_0 \times \frac{1}{2} = \frac{I_0}{2} .
   • Therefore, using the relation: K = I_0 \sin^2\left(\frac{\pi \Delta x}{\lambda}\right) , it can be inferred that at quarter wavelength, the intensity becomes \( \frac{K}{2} \).

Conclusion:

When the path difference is \( \frac{\lambda}{4} \), the intensity of light at that point is \( \frac{K}{2} \). Therefore, the correct answer is K/2.