A double slit experiment is immersed in a liquid of refractive index 1.33. Separation between the slits is 1.0 mm and he distance between slit and screen is 1.33 m. If slits are illuminated by a parallel beam of light whose wavelength is 6300 Å, then fringe width will be
Answer & explanation
Correct answer: option 3
$λt=\frac{λ_{air}}{μl}$
$β=\frac{λD}{d}$, in liquid $β=\frac{λ_1D}{d}=\frac{λ_{air}D}{μld}$
$=\frac{6300×10^{10}×1.33}{1.33×10^{-3}}=0.63mm$