Question

A light wave of wavelength \( \lambda \) is incident on a slit of width \( d \). The resulting diffraction pattern is observed on a screen at a distance \( D \). If linear width of the principal maxima is equal to the width of the slit, then the distance \( D \) is

• A. \( \frac{d}{\lambda} \)
• B. \( \frac{2\lambda}{d} \)
• C. \( \frac{d^2}{2\lambda} \)
• D. \( \frac{2\lambda^2}{d} \)

Hint:

In diffraction problems, the width of the central maxima is related to the wavelength and slit width. The larger the slit width, the smaller the diffraction pattern.

Answer 1

The Correct Option is C

Step 1: Diffraction pattern.
The angular width \( \theta \) of the first diffraction minima is given by the formula: \[ \sin \theta = \frac{\lambda}{d} \] For small angles, \( \sin \theta \approx \theta \), so the angular width of the principal maxima is \( 2\theta \).
Step 2: Linear width of the maxima.
The linear width \( W \) of the principal maxima at a distance \( D \) from the slit is: \[ W = 2 D \theta \] Using \( \theta = \frac{\lambda}{d} \), we get: \[ W = 2 D \left( \frac{\lambda}{d} \right) \] Equating \( W \) with the width of the slit \( d \), we get: \[ d = 2 D \left( \frac{\lambda}{d} \right) \] Solving for \( D \), we get: \[ D = \frac{d^2}{2\lambda} \]
Step 3: Conclusion.
The distance \( D \) is \( \frac{d^2}{2\lambda} \), so the correct answer is (C).