Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?

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(3) 22 cm. Given: n=1.55, f=20 cm, both faces of the same radius, so R₁=+R and R₂=-R. Idea: use the lens maker's formula 1/f=(n-1)(1/(R₁)-1/(R₂)). With R₁=+R and R₂=-R the bracket becomes 2/R. 1/f=(2(n-1))/R, so R=2f(n-1). R=2× 20× 0.55=22 cm Both faces are ground to this radius; the two together make the lens twice as strong as one face alone would.

Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?

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