A biconvex lens is made of glass with refractive index 1.5 and has radii of curvature 20 cm and 30 cm. If the 20 cm surface is silvered, what is the effective focal length of the mirror formed?

Explanation

Here `R_(1)=+30cm, R_(2)=-20cm,mu_(b)=1.5, mu_(b)=1`
Therefore, focal length of the lens `(1)/(f_(l))=(1.5-1)[(1)/(30)-(1)/(-20)]`
or `f_(l)=24cm`
Therefore, the 20cm side is silvered.
Focal length of the mirror,
`f_(m)=(R_(2))/(2)=(-20)/(2)=-10cm ` `[` Note the negative sign `]`
Effective focal length of the silvered lens is
`(1)/(f_(e))=(1)/(f_(m))-(2)/(f_(l)) or f_(e)=-(60)/(11) cm`
The silvered convex lens behave as a concave mirror of focal length `60//11` cm.