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, the condition for constructive interference (maxima) is given by: \[ d \sin \theta = m \lambda \] where: - \( d \) is the distance between the slits, - \( \theta \) is the angle of the maxima, - \( m \) is the order of the maxima, - \( \lambda \) is the wavelength of the light. When the experiment is conducted in a medium with refractive index \( n \), the wavelength of light in the medium changes. The new wavelength \( \lambda' \) in the medium is given by: \[ \lambda' = \frac{\lambda}{n} \] Now, in the case of the medium having refractive index \( n = \frac{4}{3} \), the wavelength \( \lambda' \) is reduced compared to the wavelength in air. The maxima in the medium will shift because the wavelength has changed. For the maximum in the medium to coincide with the 6th maximum in air, we need to adjust the order of maxima, taking into account the refractive index. Since the wavelength in the medium is smaller, the maxima will be shifted toward lower orders. Thus, the maximum in the medium corresponding to the 6th maximum in air will be the 8th maximum in the medium.