Question

The image projected by a thin equiconvex lens (n=1.70) of a frog 5.0 cm tall and 0.60 m from the

screen is to be 25 cm high. What are the necessary radii of curvature of the lens surfaces?

Answer 1

Given:
Refractive index of lens, n = 1.70
Original height of the frog, h0 = 5.0 cm
Magnified height of the frog, h1 = 25.0 cm
Image distance, v = 0.60 cm

Solution:

The object distance can be obtained from the magnification formula which is given by
m = h1/h0 = v/u

Therefore, the object distance is determined by

u = v/(h1/h0)
u = 0.60/(25.0 cm/5.0 cm)
u = 0.12 m

The necessary radius of curvature of the lens surface can be obtained from the len's formula as follows.

(n0/-u) + (n/v) = (n-n0)/R
Here, n0 - Refractive index of air = 1.00
So,
(1/-0.12 m) + (1.70/0.60 m) = (1.70-1.00)/R

Solve for R, we get
R = 0.105 m