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. `4//3`
B. `5//3`
C. `5//4`
D. `6//5`

Explanation

(b) `P_(a)=1/(f_(a))=((mu_(g)-mu_(a))/(mu_(a)))(1/(R_(1))-1/(R_(2)))`
`P_(l)=1/(t_(l))=((mu_(g)-mu_(l))/(mu_(l)))(1/(R_(1))-1/(R_(2)))`
`=5(1.5-1)/1(1/(R_(1))-1/(R_(2)))`………….(i)
`-1=(1.5-mu_(l))/(mu_(l))(1/(R_(1))-1/(R_(2)))`………(ii)
From eqs (i) and (ii) we get `mu_(l)=5/3`