Question

A Young’s double-slit experimental setup is kept in a medium of refractive index \( \frac{4}{3} \). Which maximum in this case will coincide with the 6th maximum obtained if the medium is replaced by air?

• A. 4th
• B. 6th
• C. 8th
• D. 10th

Hint:

In interference experiments, the fringe separation \( \Delta y \) is given by: \[ \Delta y = \frac{\lambda D}{d} \] where \( \lambda \) is the wavelength, \( D \) is the distance to the screen, and \( d \) is the distance between the slits. The fringe separation decreases as the refractive index increases.

Answer 1

The Correct Option is C

In Young's double-slit experiment, constructive interference (maxima) occurs when \( d \sin \theta = m \lambda \). Here, \( d \) is the slit separation, \( \theta \) is the angle of the maxima, \( m \) is the order of the maxima, and \( \lambda \) is the light's wavelength. When light passes through a medium with refractive index \( n \), its wavelength changes to \( \lambda' = \frac{\lambda}{n} \). For a medium with \( n = \frac{4}{3} \), the wavelength \( \lambda' \) decreases, causing the maxima to shift. To align the maximum in this medium with the 6th maximum in air, the order of maxima must be adjusted. Due to the reduced wavelength, maxima shift to lower orders. Therefore, the maximum in the medium corresponding to the 6th maximum in air will appear as the 8th maximum in the medium.