Question

In a single slit diffraction experiment, for slit width $ \alpha $, the width of the central maxima is $ \beta $. If we double the slit width then the corresponding width of the central maxima will be:

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

Hint:

In single slit diffraction, the width of the central maxima is inversely proportional to the slit width. Doubling the slit width will result in halving the width of the central maxima.

Answer 1

The Correct Option is C

In single slit diffraction, the angular width \( \theta \) of the central maxima is given by: \[ \sin \theta = \frac{\lambda}{\alpha} \] Where: - \( \lambda \) is the wavelength of light, - \( \alpha \) is the slit width. The width of the central maxima \( \beta \) is given by: \[ \beta = 2L \tan \theta \] Where \( L \) is the distance to the screen. Now, if we double the slit width, i.e., \( \alpha \) becomes \( 2\alpha \), the angular width \( \theta \) changes as follows: \[ \sin \theta = \frac{\lambda}{2\alpha} \] Since \( \theta \) decreases when \( \alpha \) increases, the corresponding width of the central maxima will also decrease. Hence, the new width will be: \[ \beta_{\text{new}} = \frac{\beta}{2} \] Thus, the width of the central maxima is halved when the slit width is doubled.