In Young’s double slit experiment, the fringe width is 12 mm. If the entire arrangement is placed in water of refractive index $\frac 43$, then the fringe width becomes (in mm)

Question

The exit surface of a prism with refractive index $n$ is coated with a material having refractive index $\dfrac{n}{2}$. When this prism is set for minimum angle of deviation, it exactly meets the condition of critical angle. The prism angle is ___.

Answer 1

Young's double slit experiment involves the creation of an interference pattern consisting of bright and dark fringes. The fringe width ($ \beta $) in such an experiment is given by the formula:

$\beta = \frac{\lambda D}{d}$

where $ \lambda $ is the wavelength of light used, $ D $ is the distance from the slits to the screen, and $ d $ is the distance between the slits.

When the entire setup is immersed in a medium of refractive index $ n $, the wavelength of the light changes to $ \lambda' = \frac{\lambda}{n} $. As a result, the new fringe width $ \beta' $ becomes:

$\beta' = \frac{\lambda' D}{d} = \frac{(\frac{\lambda}{n}) D}{d} = \frac{\lambda D}{nd}$

Given:

Original fringe width, $ \beta = 12 \text{ mm} $
Refractive index of water, $ n = \frac{4}{3} $

Substituting the given values into the modified formula for $ \beta' $, we have:

$\beta' = \frac{\beta}{n} = \frac{12}{\frac{4}{3}}$

Simplifying:

$\beta' = 12 \times \frac{3}{4} = 9 \text{ mm}$

Thus, the fringe width becomes 9 mm when the setup is placed in water.

Hence, the correct answer is 9 mm.

Answer 2