Answer to Question #93212 in Optics for Sridhar

Question

An object is placed 0.1 m infront of a convex lens of focal length 20 cm made of a material of refractive index 1.5. The surface of the lens away from the object is silvered. If the radius of curvature of the silvered surface is 22 cm. then the distance of the final image from the silvered surface is

Answer 1

The curved silvered surface acts like a concave lens with

$$f_m=0.5R=0.5(-22)=-11\ cm=-0.11\ m$$

$$P_m=\frac{-1}{f_m}=\frac{-1}{-0.11}=\frac{1}{0.11}D$$

$$P_l=\frac{1}{f_l}=\frac{1}{0.2}D$$

Now as in image formation, light after passing through the lens will be reflected back by curved mirror through lens again.

$$P=P_l+P_m+P_l=2\frac{1}{0.2}+\frac{1}{0.11}=\frac{210}{11}D$$

The equivalent focal length:

$$F=-\frac{1}{P}=-\frac{11}{210}\ m$$

Then,

$$\frac{1}{v}+\frac{1}{-10}=\frac{21}{110}$$

$$v=-11\ cm$$

Then the distance of the final image from the silvered surface is 11 cm.