Question

Young's double slit experiment is performed in a medium of refractive index \( 1.33 \). The maximum intensity is \( I_0 \). The intensity at a point on the screen where the path difference between the light coming out from slits is \( \lambda/4 \), is:

• A. \( 0 \)
• B. \( \frac{I_0}{2} \)
• C. \( \frac{3I_0}{8} \)
• D. \( \frac{2I_0}{3} \)

Hint:

In Young’s Double-Slit Experiment (YDSE), the intensity at a given point depends on the phase difference of the interfering waves.

Answer 1

The Correct Option is B

Step 1: {Understanding the intensity formula in YDSE}
The intensity at any point in YDSE is expressed as:\[I = I_0 \cos^2 \left(\frac{\phi}{2} \right)\]Here, \( \phi \) represents the phase difference, which is calculated using:\[\phi = \frac{2\pi}{\lambda} \times {path difference}\]Step 2: {Substituting given values}
Given a path difference \( \Delta x = \lambda/4 \), the phase difference becomes:\[\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}\]Consequently,\[I = I_0 \cos^2 \left(\frac{\pi}{4} \right) = I_0 \times \left(\frac{1}{\sqrt{2}}\right)^2\]\[I = \frac{I_0}{2}\]Therefore, the final answer is \( \frac{I_0}{2} \).