The focal length of a thin lens in vacuum is `f`. If the material of the lens has `mu = 3//2`, its focal length when immersed in water of refractive index `4//3` will be.

Options

A. f
B. `4 f//3`
C. 2 f
D. 4 f

Answer

D

Explanation

`(1)/(f_a) = ((mu_g)/(mu_a) - )((1)/(R_1) - (1)/(R_2))` …(i)
`(1)/(f_w) = ((mu_g)/(mu_w) - )((1)/(R_1) - (1)/(R_2))` ….(ii)
Dividing (i) by (ii), we get
`(f_w)/(f_a) = ((mu_g//mu_a - 1))/((mu_g//mu_w - )) = ((3//2 - 1))/((3//2)/(4//2) - 1) = (1//2)/(1//8) = 4`
`:. f_w = 4 f_a = 4 f`.