A thin lens of focal length + 12 cm is immersed in water `(mu = 1.33).` What is its new focal length ?

Explanation

We have `(1)/(f) = ((mu_2)/(mu_1) -1))((1)/(R_1) - (1)/(R_2))`
When the lens is placed in air, f = 12 cm. Thus,
`(1)/(12 cm)= (1.5 - 1) ((1)/(R_1) - (1)/(R_2))`
or,`(1)/(R_1) - (1)/(R_2) = (1)/(6 cm)`
If the focal length becomes f' when placed in water, `(1)/(f')= ((1.5)/(1.33) -1) ((1)/(R_1) - (1)/(R_2))`
`=(1)/(8)xx(1)/(6cm) = (1)/(48cm) or f' = 48 cm`