Question

In Young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes :

• A. double
• B. half
• C. four times
• D. one-fourth

Answer 1

The Correct Option is C

To solve this problem, we need to understand the formula for the fringe width in Young's double-slit experiment. The formula for fringe width \( \beta \) is given by:

\(\beta = \frac{\lambda D}{d}\)

where:

• \( \beta \) is the fringe width.
• \( \lambda \) is the wavelength of light used.
• \( D \) is the distance from the slits to the screen.
• \( d \) is the distance between the two slits (slit separation).

In the question, two changes are made to the setup:

1. The separation between the coherent sources (\( d \)) is halved.
2. The distance of the screen from the coherent sources (\( D \)) is doubled.

Let's analyze how these changes affect the fringe width \( \beta \):

• When \( d \) (slit separation) is halved: this means \( d \to \frac{d}{2} \).
• When \( D \) (distance to the screen) is doubled: this means \( D \to 2D \).

Substituting these changes into the fringe width formula, we have:

\(\beta' = \frac{\lambda (2D)}{\frac{d}{2}} = \frac{2\lambda D}{\frac{d}{2}} = \frac{4\lambda D}{d}\)

Comparing this with the original fringe width \( \beta = \frac{\lambda D}{d} \), we see that:

\(\beta' = 4\beta\)

This shows that the fringe width becomes four times the original width.

Therefore, the correct answer is: four times.