Question

A double slit interference experiment performed with a light of wavelength 600 nm forms an interference fringe pattern on a screen with 10th bright fringe having its centre at a distance of 10 mm from the central maximum. Distance of the centre of the same 10th bright fringe from the central maximum when the source of light is replaced by another source of wavelength 660 nm would be:

Hint:

In interference patterns, the fringe separation depends on the wavelength of light used. A longer wavelength results in a greater fringe separation.

Answer 1

The fringe separation in a double slit interference pattern is given by:

\[ y = \frac{\lambda D}{d} \]

where \( \lambda \) is the wavelength, \( D \) is the distance to the screen, and \( d \) is the slit separation.

Step 1: The fringe separation is directly proportional to the wavelength, so if the wavelength changes from 600 nm to 660 nm, the fringe separation will change accordingly.

Step 2: The new fringe separation will be:

\[ y_{\text{new}} = \frac{660}{600} \times 10 = 11 \, \text{mm} \]

Final Conclusion: The distance for the 10th fringe from the central maximum with the new wavelength is 11 mm, which corresponds to Option (1).