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## GR0877 Problem 15

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (E) Recall the lens-maker’s formula: $\frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$. Here $R_1$ is the radius of curvature of the lens’ surface closest to the light source and $R_2$ is the radius of curvature of the lens’ surface farthest from the light source. Sign convention: the radius of curvature is positive if the center of curvature lies to the right from the lens and is negative otherwise. Let the source of light be situated to the left from each of the lenses. Then for (A) one has $R_1 = -R$, $R_2 = R$, and $f_A \sim -R/2$. For (B): $R_1 = \infty, R2 = R$, $f_B \sim -R$. (C): $R_1 = \infty, R2 = -R$, $f_C \sim R$. For (D) and (E): $R_1 = R$, $R_2 = -R$, $f \sim R/2$. But $R_E , therefore $f_E < f_D$. Alternatively, lenses (A) and (B) are obviously diverging and from (C), (D), (E) the shortest positive focal length has the one that has the smallest radius of curvature, because it bends a set of parallel rays more than the others, hence, the focal point of this lens is the closest to the lens.

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1. 18.09.2015 at 18:25

So we have:
C) F ~ R_C
E) F ~ R_E/2

But based on the diagram, since R_C < R_E, this seems ambiguous as to which is smaller. Am I missing something?

2. 15.10.2016 at 16:13

a short focal length does not imply a small radius of curvature

• 17.10.2016 at 20:46

It does. One way to see it is to look at the lens-maker’s formula above. Another way would be to take the simplest case — plano-convex lens — and consider an incident ray parallel to its optical axis. Look where the outgoing ray intersects the optical axis (that’s the focal point) if you change the radius of curvature.

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