# Lens with a mirrored coating

## Homework Statement

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 $n$?

## Homework Equations

The previous half of the equation made us derive $\frac{n_2}{v} + \frac{n_1}{u} = \frac{n_2-n_1}{r}$ for a spherical interface of radius $r$ separating 2 different media. I don't know if this would be useful here?

Also have $\frac{2}{u} = \frac{1}{f}$

I believe we can't use the lens maker's equation in this circumstance?

## The Attempt at a Solution

Deriving the relations above.

Related Introductory Physics Homework Help News on Phys.org
mfb
Mentor
The previous half of the equation made us derive $\frac{n_2}{v} + \frac{n_1}{u} = \frac{n_2-n_1}{r}$ for a spherical interface of radius $r$ separating 2 different media. I don't know if this would be useful here?
What was the setup where you derived that?
Also have $\frac{2}{u} = \frac{1}{f}$
That does not take the mirror into account, so you have to check if you can apply this formula here.
I believe we can't use the lens maker's equation in this circumstance?
Why not?

Did you draw a sketch?

The relation was obtained by considering a spherical interface separating two different media. The image is formed at a distance $v$ in the sphere. My reasoning for not using the standard lens maker of $\frac{1}{f} = \left(\frac{n_1}{n_2} -1\right)\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$ because that assumes the light goes on to exit the lens at the other end where as here it is reflected? I did do a sketch, but it doesn't give away much...

mfb
Mentor
That equation gives a relation between the three quantities of the lens, it is independent of the light path in this specific setup.

I did do a sketch, but it doesn't give away much...
Which light paths did you draw?
Can you show the sketch here? I think you can learn a lot from the sketch.

My sketch had the lens in the mercury with rays of light being refracted as normal and then reflected and following it's path back to the source. Could it be as simple as using the derived expression in the question and the lens makers to solve for r and n?

mfb
Mentor
The lens-maker equation won't be sufficient to find both r and n, but you'll get one condition. The information about the image gives the other one.

The lens-maker equation won't tell you anything about u, but you'll get n.
True, but we can always give our answer in terms of $f$ as that is mentioned in the question?

mfb
Mentor
Sorry, updated my post to better match the homework statement. See above.

Sorry, updated my post to better match the homework statement. See above.
Ah, yes. So I'm guessing the equation calculated in the first part of the question is irrelevant in this circumstance? Otherwise I can't seem to find how the image gives any relation to $r$.

mfb
Mentor
So I'm guessing the equation calculated in the first part of the question is irrelevant in this circumstance?
I still don't know what exactly it means. What is the setup where it was calculated, what are u and v?

Every ray going from the point towards the lens/mirror combination gets reflected back in exactly the same way. That gives you a condition on the different angles and their relations.

I still don't know what exactly it means. What is the setup where it was calculated, what are u and v?

Every ray going from the point towards the lens/mirror combination gets reflected back in exactly the same way. That gives you a condition on the different angles and their relations.
My apologies. The question for the first part is as follows:
A spherical interface of radius R separates two media of refractive indices $n_1$ and $n_2$. Show that,
in the paraxial approximation, a point object on the optical axis a distance $u$ from the interface in the first medium produces a point image on the optical axis in the second medium a distance $v$ from the interface given by $\frac{n_2}{v} + \frac{n_1}{u} = \frac{n_2-n_1}{r}$