Question

The surface of a lens is convex on one side, with a radius of curvature of 50

cm, and concave on the other side, with a radius of curvature of 15 cm. If it is made of a

certain plastic with a refractive index of 1.6, find the focal length of this lens.

Answer 1

We know that

$\frac{1}{f}=(\mu-1)(\frac{1}{R_{cov}}-\frac{1}{R_{con}})$

$\mu=1.6\\R_{conv}=50cm\\R_{conc}=15cm$

Put value

$\frac{1}{f}=(1.6-1)(\frac{1}{50}-\frac{1}{{15}})$

$\frac{1}{f}=0.6\times\frac{15-50}{50\times15}$

$f=\frac{50\times15}{35\times0.6}=-35.71cm$