# Optics - Find the refractive index

1. Aug 14, 2013

### Saitama

1. The problem statement, all variables and given/known data
An object is located at a distance R from a sphere of radius R. The final image is formed at the same distance R on the other side of the sphere. Calculate the refractive index of the material of the sphere.

2. Relevant equations
Since we are dealing with refraction at spherical surfaces, the equation to be used is:
$$\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}$$

3. The attempt at a solution
I take the direction to the right as positive.
Refraction at the first surface, $\mu_2=\mu$, $\mu_1=1$ and $u=-R$:
$$\frac{\mu}{v}+\frac{1}{R}=\frac{\mu-1}{R}$$
Solving for v,
$$v=\frac{\mu R}{\mu-2}$$
I have to apply the equation for the second surface now but the problem is how do I calculate u? I don't know if the image formed by first refraction lies outside or inside the sphere.

Any help is appreciated. Thanks!

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Last edited: Aug 14, 2013
2. Aug 15, 2013

### ehild

Try ray-tracing, using the symmetry of geometry.

ehild

3. Aug 15, 2013

### Saitama

I really have no idea what you mean. Please elaborate.

4. Aug 15, 2013

### ehild

See picture. The blue and green rays are easy to follow.

ehild

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• ###### sphereimage.JPG
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5. Aug 15, 2013

### Saitama

Why the green ray deviates? It passes through the pole of refracting surface, it should pass without deviation, no?

6. Aug 15, 2013

### voko

Snell's law?

7. Aug 15, 2013

### Saitama

Silly on my part, sorry.

But I still don't see how to calculate u. :(

By u, I mean the distance of image formed by the first surface from the second surface.

8. Aug 15, 2013

### voko

The green ray has three segments, let's call them, left to right, A, B and C. Let the angles of the green ray with the normals at the interfaces be, left to right, a, b, c, d. What of these can you not calculate, given the size of the object is s and the radius of the sphere is R?

9. Aug 15, 2013

### Saitama

I can calculate all of them. Also, angle b is equal to angle c and angle d is equal to a. Correct?

10. Aug 15, 2013

### voko

What remains to be done, then?

Yes.

11. Aug 15, 2013

### Saitama

Find the location of image formed by the first refraction? I still have no idea about this.

12. Aug 15, 2013

### voko

I am not sure what "first refraction" is. The whole point of ray tracing is that two rays must meet in one location. And that should yield a condition on whatever ray tracing is used for.

13. Aug 15, 2013

### Saitama

I meant that the image formed by the refraction through the spherical surface nearest to object.
:uhh:.....

14. Aug 15, 2013

### voko

Think what is required for the two rays to meet at the image.

15. Aug 15, 2013

### Saitama

They should not diverge? :tongue2:

I am still clueless.

16. Aug 15, 2013

### voko

The two rays must meet at the specified distance behind the sphere.

17. Aug 15, 2013

### ehild

Why do you want to find the first image when you know the position of the final image? And you also know its height. And you know Snell's Law. And you can apply the small-angle approximation.

By the way, if you want to know the position of the first image at any cost: The final image is real. If the image formed by a concave surface is real, is the object real or virtual?:tongue2:

ehild

18. Aug 15, 2013

### ehild

And do not forget that the refractive index is sin(a)/sin(b), a/b at small angles.

ehild

19. Aug 15, 2013

### Saitama

Why not?

I think about the problem this way. Since the rays refract twice from the spherical surface, I have to apply the equation I posted twice. First, I apply it for the refraction at the first spherical surface (near to the object) and find the location of image formed. Now the rays refract at the second surface (surface near the final image). I will have to calculate the distance of image formed by the first surface from the second surface. The first image formed acts as the object for the second surface. I again apply the same equation for the second surface and calculate $\mu$.

Am I thinking along the wrong lines?
Object is real.

20. Aug 15, 2013

### ehild

Snell's Law is more basic than the equation you use to find the images. It can happen that the image distance is longer than 2R: the first image is behind the sphere. (It is not there really, only virtually.) The first image is object for the second surface. Normally, the object should be in front of the surface, but what is the object distance for the second spherical surface if the object is behind it?

The problem asks the refractive index. Can you determine it from the data given?

ehild