A thin glass (refractive index 1.5) lens has optical power of `-5D` in air. Its optical power in a liquid medium with refractive index 1.6 will be

Options

A. `-1 D`
B. `1 D`
C. `25 D`
D. `-25 D`

Answer

B

Explanation

(a) `(1)/(f_(a)) = (mu - 1) ((1)/(R_(1)) - (1)/(R_(2)))`
`= (1.5 - 1) ((1)/(R_(1)) = (1)/(R_(2)))` and `(1)/(f_(m)) = ((mu_(0) - mu_(m))/(mu_(m))) ((1)/(R_(1)) - (1)/(R_(2)))`
`(1)/(f_(m)) = ((1.5)/(1.5) - 1) ((1)/(R_(1)) - (1)/(R_(2)))`
Thus, `(f_(m))/(f_(a)) = ((1.5 - 1))/((1.5)/(1.6) - 1) = -8`
`f_(m) = - 8 xx f_(a)`
`= - 8 xx (-1)/(5)` (`:' f_(a) = (1)/(p) = -(1)/(5) m`)
`1.6 m`
`implies P_(m) = (mu)/(f_(m)) = (1.6)/(1.6) = 1 D`