Question:

The fringe width for red colour as compare to that for violet colour is approximately

Updated On: Apr 1, 2025
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The Correct Option is D

Solution and Explanation

Fringe Width in Double-Slit Interference

The fringe width in a double-slit interference pattern is given by the formula:

\[ w = \frac{\lambda L}{d} \]

Here, the wavelength \(\lambda\) is directly proportional to the fringe width \(w\).

- Red light has a wavelength of approximately 700 nm.
- Violet light has a wavelength of approximately 400 nm.

To compare the fringe widths:

\[ \frac{w_r}{w_v} = \frac{\lambda_r}{\lambda_v} \]

Substituting values:

\[ \frac{w_r}{w_v} \approx \frac{700}{400} \approx 1.75 \]

This means the fringe width for red light is about 1.75 times that of violet light.

Since none of the given answer choices (A, B, C, D) match this exactly, the closest option is (D) "double", suggesting that the fringe width of red light is nearly twice that of violet light.

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Concepts Used:

Wave interference

What is Interference?

When two or more waves meet each other then interference happens . Interference is a phenomenon in which 2 or more waves superpose to form a resultant wave of greater, lower or the same amplitude.

There are two types of wave interference:

The principle of superposition of waves refers that when two or more waves of the same type are incident on the same point, the resultant amplitude at that point is equal to the vector sum of the amplitudes of the individual waves. If the crest of a wave meets the crest of another wave of the same frequency at the same point,  sum of individual amplitudes is called as constructive interference.The destructive interference occurs when the maxima of the two waves are at 180 degrees out of phase and a positive displacement of one wave is cancelled exactly by a negative displacement of the other wave.