Optics - Find the refractive index

[Saitama]

Homework Statement

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.

Homework Equations

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

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!

 

[ehild]

Try ray-tracing, using the symmetry of geometry.

ehild

 

[Saitama]

Try ray-tracing, using the symmetry of geometry.

ehild

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

 

[ehild]

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


ehild

 

[Saitama]

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


ehild

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

 

[voko]

Snell's law?

 

[Saitama]

Snell's law?

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.

 

[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?

 

[Saitama]

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?

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

 

[voko]

I can calculate all of them.

What remains to be done, then?

Also, angle b is equal to angle c and angle d is equal to a. Correct?

Yes.

 

[Saitama]

What remains to be done, then?

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

 

[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.

 

[Saitama]

I am not sure what "first refraction" is.

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

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.

:uhh:...

 

[voko]

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

 

[Saitama]

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

They should not diverge? :tongue2:

I am still clueless.

 

[voko]

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

 

[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

 

[ehild]

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

ehild

 

[Saitama]

Why do you want to find the first image when you know the position of the final image?

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?

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

Object is real.

 

[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

 

[Saitama]

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 distance is $\frac{\mu R}{\mu-2}-2R=\frac{(4-\mu)R}{\mu-2}$.
I choose the direction to the right to be positive. Applying the equation to second spherical surface,
$$\frac{1}{R}-\cfrac{\mu}{\cfrac{(4-\mu)R}{\mu-2}}=\frac{1-\mu}{-R}$$
Solving this gives $0=-4$.

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

No idea but let me try making the equations.

Applying Snell's law on the first surface, $\sin(a)=\mu \sin(b)$. From small angle approximation, $\sin(a)=a=\tan(a)=s/R$, $\Rightarrow s/R=\mu b \Rightarrow b=\frac{s}{R\mu}$. Angle subtended by the object at center is s/(2R). I don't see a way ahead.

 

[ehild]

$$\frac{1}{R}-\cfrac{\mu}{\cfrac{(4-\mu)R}{\mu-2}}=\frac{1-\mu}{-R}$$
Solving this gives $0=-4$.

Your solution is wrong. Check.


How are the red and green angles related?

 

[Saitama]

How are the red and green angles related?

Are you sure you ask me those two red angles? It looks like you marked the wrong angles. I am sorry if I have missed something.

I assume you mean to ask me this:
$$\text{Red angle}=\frac{\text{Green angle}}{2}$$

 

[ehild]

Are you sure you ask me those two red angles? It looks like you marked the wrong angles. I am sorry if I have missed something.

I assume you mean to ask me this:
$$\text{Red angle}=\frac{\text{Green angle}}{2}$$

No. I asked how a red angle is related to a green one. And their ratio is not 2.

 

[Saitama]

I asked how a red angle is related to a green one. And their ratio is not 2.

Yes, you are right, their ratio is not 2. Sorry.

Red angle, y=s/(2R). Green angle, x=s/(Rμ)=2y/μ. Looks good?

 

[ehild]

The red angle is equal to the green angle. Do you see it? So s/(2R)=s/(Rμ). So what is the refractive index?


ehild

 

[Saitama]

The red angle is equal to the green angle. Do you see it?

No, do you mean that the green and blue ray are parallel? If so, how do you conclude that?

 

[ehild]

The part of the green ray inside the circle is parallel with the blue line which goes across the centre of the circle.
See figure.

 

[Saitama]

The part of the green ray inside the circle is parallel with the blue line which goes across the centre of the circle.
See figure.

Thanks, this gives $\mu=2$.

Can you tell me how you thought of applying the Snell's law and apply the small-angle approximation? The question says nothing about the approximation so how am I even supposed to think about that?

 

[ehild]

The formula you used to find the image distance is derived from Snell's Law, using small-angle approximation. Was not it shown to you?
Now, that you know the refractive index, you find the first image at infinity, beyond the sphere. And it is a virtual object for the second surface. Quite confusing is it not? But solving the problem with Snell's Law no such problem arises, everything is clear.
By the way, you get the same refractive index with your method if you solve the last equation correctly. ehild

 

[Saitama]

The formula you used to find the image distance is derived from Snell's Law, using small-angle approximation. Was not it shown to you?

Yes, it was shown to me, I forgot about it. Thanks a lot ehild! :)

By the way, you get the same refractive index with your method if you solve the last equation correctly.

I still don't get the correct answer with that equation. I always end up with 0=-4. I feel there is some sign error in my last equation but I can't spot it.

 

[ehild]

$$\frac{1}{R}-\cfrac{\mu}{\cfrac{(4-\mu)R}{\mu-2}}=\frac{1-\mu}{-R}\rightarrow$$
$$1-\frac{\mu(\mu-2)}{4-\mu}=-1+\mu\rightarrow 2-\mu+\frac{\mu(2-\mu)}{4-\mu}=0 \rightarrow \left(2-\mu\right) \left(1+\frac{\mu}{4-\mu}\right)=0$$

2-μ can be factored out, and you get a product equal to zero. One of the factors must be zero. The second one can not. Never "simplify" an equation dividing by a term that contains the unknown!

 

[Saitama]

$$\frac{1}{R}-\cfrac{\mu}{\cfrac{(4-\mu)R}{\mu-2}}=\frac{1-\mu}{-R}\rightarrow$$
$$1-\frac{\mu(\mu-2)}{4-\mu}=-1+\mu\rightarrow 2-\mu+\frac{\mu(2-\mu)}{4-\mu}=0 \rightarrow \left(2-\mu\right) \left(1+\frac{\mu}{4-\mu}\right)=0$$

2-μ can be factored out, and you get a product equal to zero. One of the factors must be zero. The second one can not. Never "simplify" an equation dividing by a term that contains the unknown!

Thanks a lot ehild! I need to be more careful while solving the equations. Thanks. :)