Question

In YDS, the width of the fringes obtained from a light of wavelength 500 nm is 3.6 mm. What is the fringe width id the apparatus is immersed in a liquid of refractive index 1.2?

(a) 2 mm

(b) 2.6 mm

(c) 3 mm

(d) 3.2 mm

Answer 1

The correct answer is (c) 3 mm

The best explanation: AS we know, Fringe width, β = λD/d, where D is the distance between the slits and the screen and d is the distance between the slits.

Now, β’ = λ’D/d

β’/ β = λ’/λ

 β’ = β/μ

     = 3.6 mm/1.2

     = 3 mm.