A convex lens of focal length 0.2 m and made of glass `9mu=1.50)` is immersed in water `9mu=1.33)`.Find the chajnge in the focal length of the lens.

Explanation

`f_("air")=0.2m=20cm`
`""^(a)mu_(g)=1.5` and `""^(a)mu_(w)=1.33`
`1/(f_("air"))=(""^(a)mu_(g)-1)(1/(R_(1))-1/(R_(2)))`
`1/20=(1.50-1)(1/(R_(1))-1/(R_(2)))`
`1/20=0.5(1/(R_(1))-1/(R_(2)))`
`implies1/(R_(1))-1/(R_(2))=1/(20xx0.5)=1/10`
`implies1/(R_(1))-1/(R_(2))=1/(20xx0.5)=1/10`
Let `f_(w)=` Focal lenght of the lens is water
Now `1/(f_(w))=(""^(w)mu_(g)-1)(1/(R_(1))-1/(R_(2)))`
`=((""^(a)mu_(g))/(""^(a)mu_(g))-1)(1/(R_(1))-1/(R_(2)))`
`1/(f_(w))=(1.5/1.33-1)(1/10)`
`=((1.5-1.33)/1.33)1/10`
`1/(f_(w))=0.17/(1.33xx10)`
`f_(w)=13.3/0.17=78.12cm`
Change in focal length `=f_(w)-f_("air")`
`=78.12-20`
`=58.12cm`
`=0.5812cm`
`=0.58m`