A glass `(mu=1.5)` convex lens of focl length 40 cm is placed in water `(mu=1.3)` .What will be its new focal length?

Explanation

`f_("air")=40cm`
`""^(a)mu_(g)=1.5` and `""^(a)mu_(w)=1.3`
`1/(f_("air"))=(""^(a)mu_(g)-1)(1/(R_(1))-1/(R_(2)))`
`1/40=(1.5-1)(1/(R_(1))-1/(R_(2)))`
`1/(R_(1))-1/(R_(2))=1/(40xx0.5)=1/20`
Let `f_(w)=` Focal length of lens in water
`1/(f_(w))=(""^(w)mu_(g)-1)(1/(R_(1))-1/(R_(2)))`
`=((""^(a)mu_(g))/(""^(a)mu_(w))-1)(1/20)`
`1/(f_(w))=(1.5/13-1)(1/20)`
`=((1.5-1.3)/1.3)1/20`
`1/(f_(w))=0.2/1.3xx1/20=1/130`
`f_(w)=130cm`