Biconvex Lens Floating on Mercury

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1. Apr 14, 2017

Marcus95

1. The problem statement, all variables and given/known data
A thin bi-convex lens with refractive index n has spherical surfaces with equal radii of curvature r and measured focal length f. The lens floats horizontally on the surface of liquid mercury so that its lower surface effectively becomes a spherical mirror. A point object on its optical axis is then found to coincide with its image when it is at a height u above the lens. Determine r and n.

2. Relevant equations
Thin lens equation: $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$
Refraction at Spherical Surface: $\frac{n_1}{u}+\frac{n_2}{v} = \frac{n_2-n_1}{R}$
Lensmakers Equation: $\frac{1}{f} = (\frac{n_2}{n_1}-1)(\frac{1}{R_1}+\frac{1}{R_2})$

3. The attempt at a solution
I am very uncertain on how to handel the spherical mirror at the back of the lens and I would like to think through the problem before beginning with calculations.
Assuming that f is known, we may use the lens makers equation to express either r or n in the other two variables.
However, I don't know how to use the other information. Had it been a straigth mirror at the other end, I think we migth be able to say that the image is simply reflected and $u = v$, but I am uncertain about this as well.
My other problem is, that if the ligth never exits the second face of the lens, maybe it cannot be viewed as a full lens? Isn't it then only a refraction at one sperical interface? Or does the spherical mirror compendate for this?

Very Thankful for explanations! :)

2. Apr 14, 2017

TSny

Yes.

I'm not quite following you here.

Yes, refraction only occurs at the upper surface of the lens. (How many times does a light ray refract through this surface?) The lower surface of the lens acts like a curved mirror.

I think the problem will simplify for you if you draw a ray diagram for a ray leaving the point object and eventually returning to the same point.

Last edited: Apr 14, 2017
3. Apr 14, 2017

Drakkith

Staff Emeritus
Hmmm. Well, you can try treating the mirror as a special refractive surface where n2 = -n1 (n2 being the index after reflection and n1 being the index prior to reflection).
You can also try Gaussian Reduction if you know how to do that.

4. Apr 18, 2017

Marcus95

The ligth ray refracts twice at that surface. Had there been a straigth mirror behind the first reflecting surface, the ray would have simply been reversed (?) and refracted again. Hence, because we know that the two sides of the lens are of equal radii we could conclude that the distance to the image is of the same magnitude as would have been if there was no mercury, but reversed, ie: $u = -v$. Now if there is a spherical mirror, I am not sure what happens. I have tried sketching ray diagrams, but the path of the ray seems to depend on n and r so I am not sure what can be said.

5. Apr 18, 2017

Marcus95

Do you mean Gaussian elimination for solving linear equation systems or is this something completely else?

6. Apr 18, 2017

TSny

A ray that leaves the object, refracts at the upper surface, and then reflects at the lower surface must return to the position of the object. What is the condition on the angle of incidence of the reflection at the lower surface so that the ray retraces its path back to the object?

7. Apr 18, 2017

Drakkith

Staff Emeritus
Something else. If you don't know what it is, don't worry about it.

8. Apr 19, 2017

Marcus95

Oh, it has to be perpendicular?

9. Apr 19, 2017

TSny

Yes. Can you use that information to tell you where the image is located for the first refraction at the upper surface?

10. Apr 22, 2017

Marcus95

I have tried to sketch the situation, but to me it seems to depend on the width of the lens where the ray is hitting the second surface perpendiculary?

11. Apr 22, 2017

TSny

The figure shows a couple of rays that leave the object, refract at the upper surface, and then meet the lower reflective surface. The red dotted lines are extensions of rays $ab$ and $cd$ back to the optic axis. So, point $I$ represents the position of the virtual image of the refraction at the upper surface.

You have correctly noted that rays $ab$ and $cd$ meet the lower surface perpendicularly to the lower surface. Use this information to deduce the image distance for $I$ in terms of the radius of curvature $r$ of the surfaces of the lens.

12. Apr 23, 2017

Marcus95

Thank you so much for the picture! I now understand that the image and object distance in order to coincide have to be equal to r, ie $r = u$ if the lens is assumed to be thin. By using the lensmakers equation we can then find $n = \frac{u}{2f} + 1$. Is that rigth?

Last edited: Apr 23, 2017
13. Apr 23, 2017

Marcus95

Or wait, wouldn't $r = u$ imply that there is no refraction?

14. Apr 23, 2017

TSny

Why do you say r = u?

The image distance must equal the object distance for the total system. But, for the refraction at the upper surface, the image distance for this refraction does not have to equal the object distance. But you should be able to deduce the image distance for this refraction using the figure.

Last edited: Apr 23, 2017