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

Explanation

From Lens maker's formula we have,
`1/(f_(a))=(mu-1)(1/R_(1)-1/R_(2))`
It is given that`mu`=1.5
`therefore1/(f_(a))=(1.5-1)(1/R_(1)-1/R_(2))`
and `1/(F_(m))=((mu_(g)-mu_(m))/(mu_(m)))(1/R_(1)-1/R_(2))`
`rArr1/(f_(m))=(1.5/1.6-1)(1/R_(1)-1/R_(2))`
Thus, `(f_(m))/(f_(a))=((1.5-1))/((1.5/1.6-1))=-8`
`rArrf_(m)=-8xxf_(a)`
`=-8xx(-1)/5=1.6 m` (`becausef_(a)=1/P=-1/5m`)
`therefore` Power of the lens, `P_(m)=mu/(f_(m))=1.6/1.6=1 D`