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

Options

A. `3/2`
B. `4/3`
C. `5/3`
D. 2

Answer

C

Explanation

(c ) When the lens in air, we have
`P_(a)=1/(f_(a))=(mu_(g)-mu_(a))/(mu_(a))[1/R_(1)-1/R_(2)]`
When the lens is in liquid, we have
`P_(1)=1/(f_(1))=(mu_(g)-mu_(l))/(mu_(l))[1/R_(1)-1/R_(2)]`
Given, `P_(a)=5,P_(l)=-1,mu_(a)=1 and mu_(g)=1.5`
On solving we get `mu_(l)=5/3`