A thin convergent glass lens `(mu_g=1.5)` has a power of `+5.0D.` When this lens is immersed in a liquid of refractive index `mu_1,` it acts as a divergent lens of focal length `100 cm.` The value of `mu_1` is

Explanation

Here, Power of glass lens in air `P = 5` dioptre
`:.` Focal length of lens in air,
`f_a = (100)/(P) = (100)/(5) = 20 cm`
Focal length of lens in liquid,
`f_l = -1 m = - 100 cm`
`mu_l = ?, mu_g = 3//2`
From lens maker's formula,
`(1)/(f_a)=((mu_g)/(mu_a)- 1)((1)/(R_1) -(1)/(R_2))`
`(1)/(20)=((3)/(2) - 1)((1)/(R_1) - (1)/(R_2)) = (1)/(2)((1)/(R_1) - (1)/(R_2))`
or `(1)/(R_1) - (1)/(R_2) = (1)/(10)`....(i)
Again, `(1)/(f_l)=((mu_g)/(mu_l)-1)((1)/(R_1) - (1)/(R_2))`
using (i), `(1)/(-100)=((3)/(2 mu_l) - 1) xx (1)/(10)`,
which gives `mu_l = 5//3`.