Question

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

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

Hint:

When the path difference in Young's double slit experiment is fractional of the wavelength, the intensity can be calculated by applying the trigonometric identity for interference.

Answer 1

The Correct Option is A

Step 1: In Young's double slit experiment, the intensity of light at a point is given by: \[ I = I_0 \cos^2 \left( \frac{\pi \Delta x}{\lambda} \right) \] where \( \Delta x \) is the path difference, and \( \lambda \) is the wavelength. If the path difference is \( \frac{\lambda}{3} \), the intensity can be calculated by substituting the value of the path difference: \[ I = I_0 \cos^2 \left( \frac{\pi \times \frac{\lambda}{3}}{\lambda} \right) = I_0 \cos^2 \left( \frac{\pi}{3} \right) = I_0 \left( \frac{1}{2} \right)^2 = \frac{I_0}{4} \] Thus, the intensity at the point with path difference \( \frac{\lambda}{3} \) is \( \boxed{\frac{I}{4}} \). \bigskip