Question

In a Young's double slit experiment, the intensity of light at a point on the screen where the path difference between the interfering waves is $\lambda$ (where $\lambda$ is the wavelength of light used) is $1$. The intensity at a point where the path difference is $\frac{\lambda}{4}$ will be (assume two waves have the same amplitude)

• A. zero
• B. 1
• C. 1/2
• D. 1/4

Answer 1

The Correct Option is C

Given:
In a Young's double slit experiment, the intensity at a point on the screen where the path difference is \( \lambda \) is \( I = 1 \). We are asked to find the intensity at a point where the path difference is \( \frac{\lambda}{4} \).

Step 1: Intensity Formula
The intensity of the light at any point on the screen is given by the formula: \[ I = I_{\text{max}} \cos^2 \left( \frac{\pi \Delta x}{\lambda} \right) \] where \( I_{\text{max}} \) is the maximum intensity, and \( \Delta x \) is the path difference between the two interfering waves. 

Step 2: Path Difference \( \lambda \)
When the path difference is \( \lambda \), the intensity is at its maximum, i.e., \( I_{\text{max}} = 1 \). Therefore, the intensity at this point is: \[ I = \cos^2 \left( \frac{\pi \lambda}{\lambda} \right) = \cos^2 (\pi) = 1 \] 

Step 3: Path Difference \( \frac{\lambda}{4} \)
When the path difference is \( \frac{\lambda}{4} \), we substitute into the intensity formula: \[ I = \cos^2 \left( \frac{\pi \cdot \frac{\lambda}{4}}{\lambda} \right) = \cos^2 \left( \frac{\pi}{4} \right) \] Since \( \cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \), we have: \[ I = \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2} \] 

Answer: The intensity at a point where the path difference is \( \frac{\lambda}{4} \) is \( \frac{1}{2} \).