Question

What was the focal length of the lens when the radius of curvature was .70m and the index of refraction was 1.8?

Answer 1

We know the lens maker formula for thin lens is

`color (magenta)(1/f=(mu-1)(1/R_1-1/R_2))`

where

`fto` focal length.

`muto` refractive index of the medium of the lens w.r. to air.

`R_1andR_2` are two radii of curvature of two surfaces of the lens.

When the lens is convex

then `R_1=0.7m and R_2=-0.7m`

And `mu=1.8`

So by lens maker formula we get

`color (blue)(1/(f_"convex") =(mu-1)(1/R_1-1/R_2))`

`color (blue)(=>1/(f_"convex") =(1.8-1)(1/0.7+1/0.7))`

`color (blue)(=>1/(f_"convex") =0.8xx2/0.7`

`color (blue)(=>f_"convex" =7/16=0.4375m`

When the lens is concave

then `R_1=-0.7m and R_2=0.7m`

And `mu=1.8`

So by lens maker formula we get

`color (green)(1/(f_"concave") =(mu-1)(1/R_1-1/R_2))`

`color (green)(=>1/(f_"concave") =(1.8-1)(-1/0.7-1/0.7))`

`color (green)(=>1/(f_"concave") =-0.8xx2/0.7`

`color (green)(=>f_"concave" =-7/16=-0.4375m`