Question

In Young's double slit experiment with light of wavelength \( \lambda \), the intensity of light at a point on the screen where the path difference becomes \( \frac{\lambda}{3} \) is (I is intensity of the central bright fringe)

• A. I
• B. \( \frac{I}{2} \)
• C. \( \frac{I}{3} \)
• D. \( \frac{I}{4} \)

Hint:

Remember the relation \( \Delta \phi = \frac{2\pi}{\lambda} \Delta x \). If the path difference is \( \lambda/n \), the phase difference is \( 2\pi/n \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:

The intensity of light in Young's Double Slit Experiment (YDSE) is given by the formula \( I' = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right) \), where \( I_{\text{max}} \) is the maximum intensity (central bright fringe) and \( \phi \) is the phase difference corresponding to the path difference \( \Delta x \).
Step 2: Key Formula or Approach:

1. Phase difference \( \phi = \frac{2\pi}{\lambda} \Delta x \). 2. Intensity relation: \( I' = I \cos^2\left(\frac{\phi}{2}\right) \).
Step 3: Detailed Explanation:

Given path difference \( \Delta x = \frac{\lambda}{3} \). Calculate the phase difference \( \phi \): \[ \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3} = 120^\circ \] Now, calculate the intensity \( I' \): \[ I' = I \cos^2\left(\frac{\phi}{2}\right) = I \cos^2\left(\frac{120^\circ}{2}\right) = I \cos^2(60^\circ) \] We know that \( \cos(60^\circ) = \frac{1}{2} \). \[ I' = I \left(\frac{1}{2}\right)^2 = \frac{I}{4} \]
Step 4: Final Answer:

The intensity at that point is \( \frac{I}{4} \).