Question

How is the converging/diverging power of a lens related to its radius of curvature?

Answer 1

According to the lens maker's formula,

1f=(n2n1−1)(1R1−1R2)1f=(n2n1−1)(1R1−1R2),

Here, n2n2 is the refractive index of the lens and n1n1 is the refractive index of the medium from which the light is incident on the lens. The medium surrounding the lens must be the same for this formula to work.

R1R1 and R2R2 are the radius of curvature of the curved surfaces of the lens from which the light enter the lens and emerges out of the lens respectively.

Sign Conventions play an important role here.

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For a convex lens, R1R1 is +ve (if light is incident from the left) and R2R2 is -ve. In the above image, as you can see, the radius of the curvature of the convex lens' left curved surface (R1)(R1) lies in the direction of the incident ray, so it is +ve and the radius of curvature of the other surface (R2)(R2) points the direction opposite to the direction of incident ray, so its -ve.

So after applying the sign conventions for a convex lens, as explained above, we get the following formula,

1f=(n2n1−1)(1R1+1R2)1f=(n2n1−1)(1R1+1R2)

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