A double-slit apparatus is immersed in a liquid of refractive index $\mu=\frac{4}{3}$. It has slit separation of $10^{-3} m$ and distance between the plane of slits and screen is 133 cm. The slits are illuminated by a parallel beam of light of wavelength 630 nm in air. What is the fringe width?

0.63 mm

The correct answer is Option (3) → 0.63 mm

The fringe width (β) in a double slit experiment is given by -

$β=\frac{λ'D}{d}$

where,

$λ'$ = Wavelength of light in the medium

D = 1.33 m [Distance between slit and screen]

d, Separation between slits = $10^{-3}m$

and,

$λ'=\frac{λ}{μ}$ 

$λ =630nm$ wavelength in air [given]

$λ'=\frac{630×10^{-9}}{\frac{4}{3}}=472.5×10^{-9}m$

$∴β=\frac{630×10^{-9}×1.33}{10^{-3}}=628.425×10^{-6}m$

$≈0.63mm$