In a Young’s double-slit experiment, the fringe width is found to be ẞ. If the entire apparatus is immersed in a liquid of refractive index µ, the new fringe width will be?

Q) In a Young’s double-slit experiment, the fringe width is found to be ẞ. If the entire apparatus is immersed in a liquid of refractive index µ, the new fringe width will be?
a) β
b) μβ
c) β/μ
d) β/μ^2

Correct Answer: c) β/μ

In a Young’s double-slit experiment, the fringe width $\beta$β is given by:$$\beta = \frac{\lambda D}{d}$$

where:

$d$ = distance between the two slits

$\lambda$ = wavelength of light in the medium

$D$ = distance between slits and screen

When the apparatus is immersed in a liquid of refractive index $\mu$:

The wavelength in the medium becomes:$$\lambda’ = \frac{\lambda}{\mu}$$λ′=μλ​

Since $D$D and $d$d remain unchanged, the new fringe width $\beta’$β′ is:$$\beta’ = \frac{\lambda’ D}{d} = \frac{(\lambda/\mu) D}{d} = \frac{1}{\mu} \cdot \frac{\lambda D}{d}$$ $$\beta’ = \frac{\beta}{\mu}$$