A thin lens made of glass of refractive index `muu = 1.5` has a focal length equal to 12 cm in air. It is now immersed in water (`mu=4/3`). Its new focal length is

Options

A. 48 cm
B. 36 cm
C. 24 cm
D. 12 cm

Answer

A

Explanation

(a) Focal length in air is given by,
`1/(f_(a))=(""_(a)mu_(g)-1)(1/R_(1)-1/R_(2))`
The focal length of lens immersed in water is given by,
`1/(f_(l))=(""_(l)mu_(g)-1)(1/R_(1)-1/R_(2))`
where `R_(1),R_(2)` are radii of curvature of the two surfaces of lens and `""_(l)mu_(g)` is refractive index of glass with respect to liquid.
Also, `""_(l)mu_(g)=(""_(mu_(g))/(""mu_(l))`
Given, `""_(a)n_(g)=1.5,f_(a)=12 cm and ""_(a)mu_(l)=4//3`
`therefore (f_(l))/(f_(a))=((""_(a)mu_(g)-1)/(""_(l)mu_(g)-1))`
`(f_(l))/(12)=((1.5-1)/(1.5/(4//3)-1))=(0.5xx4)/0.5`
`rArr f_(l)=4xx12=48 cm`