Q. Young's double slit interference apparatus is immersed in a liquid of refractive index $1.44$. It has slit separation of $1.5$ mm. The slits are illuminated by a parallel beam of light whose wavelength in air is $690$ nm. The fringe-width on a screen placed behind the plane of slits at a distance of $0.72$ m, will be :
A) 0.23 mm
B) 0.33 mm
C) 0.63 mm
D) 0.46 mm
Explanation
[Solution by Dr. Abubacker] $\beta=\frac{\lambda D}{\mu d}=\frac{690\times10^{-9}\times0.72}{1.44\times1.5\times10^{-3}}=2.3\times10^{-4}$ m $=0.23$ mm. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.