Question

In Young's double slit experiment, the resultant intensity of light at a point on the screen is \( I \) when the path difference is \( \lambda \). When the path difference is \( \frac{\lambda}{4} \), the intensity at a point will be ( \( \lambda \) = wavelength of light, \( \cos 180^\circ = -1, \cos 45^\circ = \frac{1}{\sqrt{2}} \))

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

Hint:

In interference patterns, the intensity is proportional to the square of the cosine of the phase difference.

Answer 1

The Correct Option is B

Step 1: Understanding the intensity formula.
The intensity of light in Young's double slit experiment is given by the formula: \[ I = I_0 \cos^2 \left( \frac{\pi \Delta x}{\lambda} \right) \] where \( I_0 \) is the maximum intensity, and \( \Delta x \) is the path difference.
Step 2: Using the given path difference.
For a path difference of \( \frac{\lambda}{4} \), the intensity is: \[ I = I_0 \cos^2 \left( \frac{\pi}{4} \right) = I_0 \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{I_0}{2} \] Step 3: Conclusion.
Thus, the intensity at a point will be \( \frac{I}{2} \), which corresponds to option (B).