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A biconvex lens of refractive index $n$ and radius of curvature $r$ and focal length $f$ floats horizontally on liquid mercury such that its lower surface is effectively a spherical mirror. A point object on the optical axis a distance $u$ away is then found to coincide with its image. What are $r$ and $n$?

The previous part of the question yielded a proof of : $$\frac{n_2}{v} + \frac{n_1}{u} = \frac{n_2-n_1}{r}$$

However I'm not too sure on how to progress through the latter half of the question.

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closed as off-topic by Rob Jeffries, Kyle Kanos, John Rennie, Brandon Enright, JamalS Feb 10 '15 at 8:19

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Break the problem into two parts. First consider the convex mirror. A point source at the center of curvature would be reflected onto itself. We can equally well describe the rays coming out of the mirror as parallel beams refracted by a lens with focal length $f_1= r$.

enter image description here

Now if we put a second lens in front, the combined lens will have a power given by

$$\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{f}$$

And from the lensmaker's formula we know that a plano convex lens has focal length

$$f_2=\frac{r}{n-1}$$

Combining these we get

$$u=\frac{r}{n}$$

Now we combine with the information that radius of curvature $r$ led to initial focal length $F$, which means that

$$r=2(n-1)F$$

If my back-of-a-napkin math is right, it follows that

$$n=\frac{u}{2F}-1$$

And that

$$r=u-4F$$

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