Question

The convex side of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature of 20cm. The concave surface has a radius of curvature of 60cm. What is the focal length of the lens? The convex side is silvered and placed on a horizontal surface. What is the effective focal length of the silvered lens? The concave part is filled with water with refractive index 1.33. What is the effective focal length of the combined glass and water lens? If the convex side is silvered what is the new effective focal length of the silvered compound lens?

Answer 1

`(1)/(f_(1))=(1.5-1)[(1)/(20)-(1)/(60)]=(1)/(60)`
`f_(1)=60cm`
When convex side is silvered and water is not filled:
`(1)/(F)=(2)/(f_(1))-(1)/(f_(m))=(-R_(1))/(2)=-10cm`
`(1)/(F)=(1)/(-10)-2[(1)/(60)]`
`F=(-30)/(4)=-7.5cm`
After water is filled :
Let `f_(l_(1))` is the focal length of water lens:
`(1)/(f_(l_(1)))=((4)/(3)-1)((1)/60)=(1)/180`
`(1)/(F) =(2)/(f_(l_(1)))-(2)/(f_(l_(2)))-(1)/(f_(m))`
Solving, we get `F=(-90)/(13)cm`