Question

A bi-concave lens has a radius of curvature of one side equal to 10.5cm and on the other side equal to 15.0cm.If the lens is made of crown glass(n=1.52), what is the focal length of the lens?

Answer 1

$R_1=10.5cm$

$R_2=15cm$

$n=1.52$

$\frac{1}{F}=(n-1)(\frac{1}{R_2}+\frac{1}{R_1}) \to F = \large\frac{R_2*R_1}{(n-1)(R_1+R_2)}=\frac{10.5*15}{0.52*25.5}=$ $11.88$