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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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


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


Combining these we get


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


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


And that



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