The correct option is A 1.8
When the apparatus is immersed in a liquid, λ is reduced μ(refractive index) times.
i.e. μ=λvλl
Now, β=λDd
⇒βvβl=λvλl=μ
∴βv=μ βl....(i)
Now, 6th dark band in vacuum coincides with 10th bright band in the liquid.
⇒[β6,d]v=[β10,b]l
(2×6−12)βv=10 βl
From (i), we can write,
112 μ βl=10 βl
∴μ=2011=1.81
Hence, (A) is the correct answer.
Why this question?Since the fringe width depends on the wavelength of light, \ the fringe width decreases when the whole set up \ is put in a liquid of refractive index μ.