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 given by: \[ I = I_0 \cos^2 \left(\frac{\phi}{2} \right) \] where \( \phi \) is the phase difference, given by: \[ \phi = \frac{2\pi}{\lambda} \times {path difference} \] Step 2: {Substituting given values}
For the given path difference \( \Delta x = \lambda/4 \), \[ \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2} \] Thus, \[ I = I_0 \cos^2 \left(\frac{\pi}{4} \right) = I_0 \times \left(\frac{1}{\sqrt{2}}\right)^2 \] \[ I = \frac{I_0}{2} \] Thus, the correct answer is \( \frac{I_0}{2} \).

Answer 2

Step 1: Understanding the YDSE in a Medium
Young’s Double Slit Experiment (YDSE) shows an interference pattern due to the **path difference** between the light from the two slits. Even when performed in a **medium** (with refractive index \( n \)), the interference pattern forms similarly — but the **wavelength changes** to \( \lambda_n = \frac{\lambda}{n} \). However, for intensity calculation, only the **phase difference** matters.

Step 2: Given Data
- Path difference: \( \Delta x = \frac{\lambda}{4} \)
- Refractive index \( n = 1.33 \) — this doesn't change the result for this question, since we are already given the **path difference in terms of \( \lambda \)**.
- Maximum intensity: \( I_0 \)

Step 3: Phase Difference
The phase difference \( \delta \) corresponding to path difference \( \Delta x \) is: \[ \delta = \frac{2\pi}{\lambda} \cdot \Delta x = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2} \]

Step 4: Calculating Intensity at this Phase Difference
The intensity at any point in an interference pattern is given by: \[ I = I_0 \cos^2\left(\frac{\delta}{2}\right) \] Substituting \( \delta = \frac{\pi}{2} \): \[ I = I_0 \cos^2\left(\frac{\pi}{4}\right) = I_0 \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2} \]

Step 5: Final Answer
\[ \boxed{I = \frac{I_0}{2}} \] So, the correct option is: Option 2.