Question

In the Young's double slit experiment, the 2nd bright fringe for red light coincides with the 3rd bright fringe for violet light. Then the value of 'n' is: (Given: wavelength of red light = 6300 $\text{Å}$ and wavelength of violet light = 4200 $\text{Å}$)

• A. 2
• B. 4
• C. 3
• D. 1

Hint:

In Young's double slit experiment, the fringe positions for different wavelengths can be compared using the relationship \( n \lambda_{\text{red}} = (n+1) \lambda_{\text{violet}} \).

Answer 1

The Correct Option is A

In Young's double slit experiment, the position of the \(n^{\text{th}}\) bright fringe is given by the equation: \[ x_n = \frac{n \lambda D}{d}, \] where: - \( x_n \) is the position of the \(n^{\text{th}}\) bright fringe, - \( n \) is the order number of the fringe, - \( \lambda \) is the wavelength of the light used, - \( D \) is the distance between the slits and the screen, - \( d \) is the distance between the two slits. In this problem, we are given that the \(2^{\text{nd}}\) bright fringe for red light coincides with the \(3^{\text{rd}}\) bright fringe for violet light. This means that: \[ x_2 \text{ (for red)} = x_3 \text{ (for violet)}. \] From the equation, this can be written as: \[ 2 \lambda_{\text{red}} \frac{D}{d} = 3 \lambda_{\text{violet}} \frac{D}{d}. \] Since \( \frac{D}{d} \) is common for both, we can cancel it out, and the equation becomes: \[ 2 \lambda_{\text{red}} = 3 \lambda_{\text{violet}}. \] Substitute the given values for the wavelengths: \[ 2 \times 6300 = 3 \times 4200, \] \[ 12600 = 12600. \] Thus, the value of \(n\) is 2, which corresponds to option (A).