Question

In Young's double slit experiment, the intensity on the screen where the path difference between the two interfering waves is \( \frac{\lambda}{4} \) is \( K \). What will be the intensity at the point where the path difference is \( \frac{\lambda}{4} \)?

• A. \( \frac{K}{4} \)
• B. \( \frac{K}{2} \)
• C. \( K \)
• D. Zero

Hint:

In Young's double slit experiment, the intensity depends on the cosine square of the phase difference. For path differences like \( \frac{\lambda}{4} \), you can use the cosine formula to determine the intensity.

Answer 1

The Correct Option is B

The intensity in Young's double slit experiment is given by the equation:

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

Where:
- \( I_0 \) is the maximum intensity,
- \( \Delta x \) is the path difference,
- \( \lambda \) is the wavelength.

Given that the path difference \( \Delta x = \frac{\lambda}{4} \), we can substitute this value into the intensity equation:

\[ I = I_0 \cos^2 \left( \frac{\pi \times \frac{\lambda}{4}}{\lambda} \right) \]

Simplifying:

\[ I = I_0 \cos^2 \left( \frac{\pi}{4} \right) \]

Since \( \cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \), we get:

\[ I = I_0 \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{I_0}{2} \]

Thus, the intensity is \( \frac{K}{2} \), so the correct answer is \( \frac{K}{2} \).