Question

We would like to place an object 45.0cm in front of a lens and have its image appear on a screen 90.0cm behind the lens. What must be the focal length of the appropriate positive lens? If it is a bi-convex lens of refractive index 1.5, what are the values of radii of curvature?

Answer 1

the radii of curvature is 30 cm.

Step-by-step explanation:

given,

object is place at = 45 cm

image appears at = 90 cm

focal length = ?

refractive index = 1.5

radii of curvature = ?

`(1)/(f) = (1)/(u) +(1)/(v)`

`(1)/(f) = (1)/(45) +(1)/(90)`

f = 30 cm

using lens formula

`(1)/(f) = (n-1)((1)/(R_1) -(1)/(R_2))`

`R_1 = R\ and\ R_2 = -R`

`(1)/(f) = (n-1)((1)/(R) +(1)/(R))`

`R = (n -1)\ f`

`R = 2(1.5 -1)\ 30`

R = 30 cm

hence, the radii of curvature is 30 cm.