Question

Equi-convex lenses are to be manufactured from glass with a refractive index of 1.55, with both faces having the same radius of curvature. What is the radius of curvature required if the focal length is 10 cm?

Answer 1

Refractive index of glass, μ = 1.55

Focal length of the convexo-concave lens, f = 10 cm

Radius of curvature of one face of the first convex surface = R₁

Radius of curvature of the other face of the second convex surface = −R₁

Therefore, R₁ = R and R₂ = −R

The value of R can be calculated from Lens Maker formula:

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

`1/10 = (1.55 - 1) (1/R + 1/R)`

`1/10 = 0.55 xx 2/R`

R = (0.55 × 2 × 10)

R = 11 cm

Hence, the radius of curvature of the convexo-concave is 11 cm.