Question

In Young's double slit experiment, the intensity of light at a point on the screen is \( K \) unit for path difference \( \Delta \). What would be the intensity at a point if the path difference is \( \frac{\Delta}{4} \)?

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

Hint:

In Young's double slit experiment, the intensity varies with the cosine squared of the phase difference, which is related to the path difference. Use this relationship to calculate intensity at different points.

Answer 1

The Correct Option is D

Step 1: Understanding Young's double slit interference.
In Young's double slit experiment, the intensity at a point on the screen is given by the interference of the waves from the two slits. The intensity \( I \) at a point is related to the path difference \( \Delta \) by the formula: \[ I = I_0 \cos^2 \left( \frac{\pi \Delta}{\lambda} \right) \] where \( \lambda \) is the wavelength of the light, and \( I_0 \) is the maximum intensity.
Step 2: Substituting the new path difference.
If the path difference is \( \frac{\Delta}{4} \), the intensity becomes: \[ I = I_0 \cos^2 \left( \frac{\pi \Delta / 4}{\lambda} \right) \] Using the given intensity \( K \) for the path difference \( \Delta \), we substitute and simplify to find the new intensity at \( \frac{\Delta}{4} \), which turns out to be \( \frac{K}{2} \).
Step 3: Conclusion.
Thus, the intensity at the point where the path difference is \( \frac{\Delta}{4} \) is \( \frac{K}{2} \), which corresponds to option (D).