# Homework Help: Light reflected by uniform glass sphere

1. Sep 24, 2010

### ceran

1. The problem statement, all variables and given/known data
S solid uniform glass sphere is surrounded by air. The sphere has radius R and refractive index n. As shown in the picture, a light ray travelling in air, parallel to a diameter of the sphere, enters the sphere, reflects off the far side, then exits the sphere travelling in a direction exactly opposite to its original direction.

Showing all details of your solution, find the relationship that expresses x (for 0<x<R) in terms of n and R in the situation shown in the picture.

2. Relevant equations

Snell's Law, nasin(thetaa) = nbsin(thetab)

3. The attempt at a solution

I labelled the angle of incidence and angle of refraction as a and b in the diagram.
Since n for air is one, the equation should become
sin(a) = nbsin(b)

Due to geometry, sin(a) will be equal to x/R. However, I cannot figure out an expression for sin(b) in terms of x, R, and n. I have tried for a long time, and just cannot figure out the geometry of the problem. Thanks for your help.

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2. Sep 24, 2010

### Redbelly98

Staff Emeritus
Welcome to physics forums.

Consider the triangle formed by (A) the ray's entrance point on the sphere's surface, (B) the point where the ray is reflected, and (C) the sphere's center.

Clearly the angle at (A) is _____. What are the other angles in that triangle?

3. Sep 25, 2010

### ceran

Thanks for the quick response. The angle at (A) will be the angle of refraction. The angle at the point where the light is reflected would be the angle of reflection, and the third angle would be 360 degrees minus the two angles. I still don't see how the sine of the angle of refraction could be written in terms of n,x,and R though.

Last edited: Sep 25, 2010
4. Sep 25, 2010

### Redbelly98

Staff Emeritus
Pretty much. I am guessing the "360" you wrote is a typo, since the angles in a triangle actually add up to ____.
Let's see what else we can figure out about the triangle ABC. What do you know about the lengths of any of its sides?

5. Sep 25, 2010

### ceran

Yeah, the 360 was a typo haha. The angles in a triangle add up to 180 degrees.
Well, the sides AC and BC would each be equal to R, the radius of the sphere. Side AB is unknown.

6. Sep 25, 2010

### reicher89

I'm having the exact same problem as you.
All I can come up with is x = 2*tan(theta_r)*R
I can't figure out how to have the answer in terms of R and n. Only theta_r and R

7. Sep 25, 2010

### Redbelly98

Staff Emeritus
Good, and therefore it is what type of triangle?

And, what else is true about that type of triangle?

8. Sep 25, 2010

### reicher89

it is an isosceles triangle.

9. Sep 25, 2010

### Redbelly98

Staff Emeritus
Yes, and what else (besides having 2 sides equal) is true about that type of triangle?

10. Sep 25, 2010

### ceran

For an isosceles triangle, the two angles opposite from the two equal sides would be equal. However, I still can't express sin(B) in terms of n,x, and R. I tried sine rule, but it becomes
0 = 0 if I use the equal sides, and the other side and angle are unknown. I also tried using angle B = 2A, and then using the double angle formula for sine to solve for sin(B), but then I need an expression for cos(B) as well.

11. Sep 25, 2010

### rl.bhat

The important point about this problem is, why the light reflects inside the sphere instead of refraction? Is it true for any value of x?
Here you have consider the total internal reflection and the critical angle.
As x increases, angle of incidence increases and angle of refraction also increases. At one particular value of x, the angle of refraction equals the critical angle of the medium.
Do you know the relation between the critical angle and the refractive index?

12. Sep 26, 2010

### Redbelly98

Staff Emeritus
You can use that fact to come up with another equation relating angles a and b. Just to convince myself I'm not crazy, I have gone ahead and solved the problem using that information.

That equation (the one that you must come up with, using what you know about isosceles triangles) plus Snell's law gives you two equations in the two unknowns a and b. So you can actually determine what a and b are, in terms of n.

Is that a typo? It's not really true, at least not the way you wrote it. But you should be able to come up with a similar-looking equation.

Actually, the light is both reflected and refracted inside the sphere. We are just concerning ourselves with the reflected part. Total internal reflection is irrelevant here.

13. Sep 26, 2010

### ceran

Yeah it is a typo. But A = 2B should be correct right? However, I must be doing something wrong, as my attempt at a solution is becoming very complicated, with many variables at high powers and several square roots. I have tried using both the sine law and the cosine law, but it seems to get too complicated to be correct.
I used Snell's Law and I came up with x/R = nsin(B). Also , if I call the unknown angle in the triangle angle C, I used the multiple angle formulae for sine and cosine to come up with sin(C) = sin(A) and cos(C) = cos(A). But I am finding it impossible to come up with an expression for sin(B)! The question wants me to solve for x. What other properties of isosceles triangles could I use besides the equal angles and equal sides? Could you give me another hint? This problem has been bugging me for the past several days. Thanks so much for your help so far.

14. Sep 26, 2010

### Hitsuzen

Same question.

http://img214.imageshack.us/img214/4415/phys369assign11.png [Broken]

I'm also stuck trying to isolate sin(b) or put it in terms of x,n, R.

Last edited by a moderator: May 4, 2017
15. Sep 26, 2010

### Redbelly98

Staff Emeritus
Yes, good. So to recap, we have
A = 2B
sinA = n sinB (Snell's Law)
That's 2 equations with 2 unknowns. Try solving that system, and thus find A and B, before bringing x and R into the picture. The A=2B equation suggests a double-angle formula might come in handy.

Uh, I disagree with the cos equation here. If both of those were true, you'd have A=C, which is clearly not true from the figure. (Angle A is acute, while angle C is obtuse.)

You've already figured out that A=2B, so there's no more to do on that front. See my hint above about trig double-angle formulas.
You're welcome.

Last edited: Sep 26, 2010
16. Sep 26, 2010

### ceran

Yeah I found that to be a bit fishy too ha. I must have made a mistake.
I used the two equations
A = 2B and sin(A)=nsin(B)
I first solved for B, then rewrote sin(A) as x/R.
I ended up with a solution of x=nRsin(arccos(n/2)). Would this be correct?

17. Sep 26, 2010

### Redbelly98

Staff Emeritus
Yes, it is correct.

I'll just mention it is possible to express sin(arccos(n/2)) as an algebraic expression, i.e. without the trig functions. But you have indeed found the solution.

18. Sep 26, 2010

### ceran

Thank you so much for your help. Ive been struggling with optics so far haha.

19. Sep 26, 2010

### boniant

I too was stuck on this question, and ended up with an answer of:

x = R*sqrt(n^2-4).

I started with:

sina=nsinb (a being angle of incidence, b being angle of refraction)

Then since sina = x/R
x/R = nsinb

And sinb = x/sqrt(x^2 + (2R)^2) due to Pythagoras.

So the equation then becomes:

x/Rn = x/sqrt(x^2 + (2R)^2)

sqrt(x^2 + (2R)^2) = Rn

then isolating x gives:

x = Rsqrt(n^2 - 4)

Is this right? Or did I screw up in my algebra somewhere.

EDIT:: hm I just noticed that with my equation, if n is <4 (which it is) I get an imaginary number... crap
I honestly can't find where I made a mistake though.

Last edited: Sep 26, 2010
20. Sep 26, 2010

### Hitsuzen

Where did you get $$2R$$? It should be something like $$2R-c_1$$, where $$c_1$$ is the distance from the surface of the of the sphere to the to where you drew the line down to the optical axis to make a right triangle. That or you're right and I've got the wrong triangle in my head.

http://yfrog.com/5zphys369assign12p" [Broken]

Last edited by a moderator: May 4, 2017