A convex lens of refractive index 1.5 has a focal length of 18 cm in air.Calculate the change in focal length,when it is immersed in water of refractive index `4//3`.

Explanation

`f_("air")=+20cm`
`""^(a)mu_(g)=1.5` and `""^(a)mu_(g)=4//3 `
`R_(1)=R`(say)
For plane surface of lens `R_(2)=oo`
`1/(f_("air"))=(""^(a)mu_(w)-1)(1/(R_(1))-1/(R_(2)))`
`1/20=(1.5-1)(1/R-1/(oo))=0.5(1/R)`
`1/20=0.5/R`
`impliesR=20xx0.5=10cm`
Let `f_(w)=` Focal length of the lens in water
`1/w=(""^(w)mu_(g)-1)(1/R-1/(oo))`
`1/(f_(w))=((""^(a)mu_(g))/(""^(a)mu_(w))-1)(1/10)`
`1/(f_(w))=(1.5/(4//3)-1)1/10=(4.5/4-1)1/10`
`1/(f_(w))=0.5/40`
`1/(f_(w))=40/0.5=80cm`
Change in focal length of lens `=f_(w)-f_("air")`
`=80-20=60cm`