Question

Two sources give an interference pattern observed on a screen distance D away. The fringe width is 2W. If D is doubled, the fringe width will:

• A. become W/2
• B. remains the same
• C. become W
• D. become 4W

Hint:

Fringe width is directly proportional to screen distance D.

Answer 1

The Correct Option is D

To solve this question, we need to understand the concept of fringe width in an interference pattern, which is created due to the superposition of waves from two coherent sources.

The fringe width, denoted by \(W\), in a double-slit interference pattern can be calculated using the formula:

\(W = \frac{\lambda \cdot D}{d}\)

where:

• \(\lambda\) is the wavelength of the light used,
• \(D\) is the distance between the screen and the slits,
• \(d\) is the separation between the slits.

Initially, we are given that the fringe width \(= 2W\). According to the question, the distance \(D\) is doubled. Let's see how this affects the fringe width.

New fringe width:

\(W_{\text{new}} = \frac{\lambda \cdot (2D)}{d} = 2 \cdot \left(\frac{\lambda \cdot D}{d}\right) = 2W\)

Since the initial fringe width was \(2W\) with distance \(D\), doubling the distance \(D\) will result in the new fringe width:

\(W_{\text{new}} = 2 \cdot 2W = 4W\)

Therefore, the correct answer is that the fringe width will become 4W when the distance \(D\) is doubled.