# Homework Help: Ray optics - Reflection from sphere

1. Jul 3, 2013

### Saitama

1. The problem statement, all variables and given/known data
A light ray parallel to the x-axis strikes the outer reflecting surface of a sphere at a point (2,2,0). Its center is at the point (0,0,-1). The unit vector along the direction of reflected ray is $x\hat{i}+y\hat{j}+z\hat{k}$. Find the value of $yz/x^2$.

2. Relevant equations

3. The attempt at a solution
The vector for the light ray travelling parallel to x-axis is $-2\hat{i}$. The incident ray and the reflected ray make equal angle with the normal passing through (2,2,0). The vector associated with the normal at (2,2,0) is $-2\hat{i}-2\hat{j}-\hat{k}$.
Dot product of the incident ray and normal:
$$4=2\times 3 \cos \theta \Rightarrow \cos\theta = \frac{2}{3}$$
where $\theta$ is the angle between the normal and incident ray.
Dot product of reflected and incident ray:
$$-2x=2\cos 2\theta \Rightarrow x=\frac{1}{9}$$
Dot product of normal and reflected ray:
$$-2x-2y-z=3\cos \theta \Rightarrow 2y+z=\frac{-20}{9}$$
I still need one more equation.

Any help is appreciated. Thanks!

2. Jul 3, 2013

### voko

Can you think of any plane where the reflected ray has to be?

3. Jul 3, 2013

### Saitama

Sorry, I am really clueless on this.

4. Jul 3, 2013

### voko

Project the picture onto the tangential plane. The projection of the normal is a single point. The projection of the incident ray is a straight line that ends at that point. The projection of the reflected ray is a straight line that starts at that point. So these two projections are connected at that point, and what is the angle between them?

5. Jul 3, 2013

### Saitama

The tangent plane at (2,2,0) is $2x+2y-z=8$. Projection of vector $-2\hat{i}$ on this plane is $\displaystyle \frac{-10\hat{i}+8\hat{j}+4\hat{k}}{9}$. Correct? (I would like to confirm if what I am doing is right. I am not familiar with solid geometry, I found the projection and the plane by looking up the information online.)

6. Jul 3, 2013

### voko

I did not mean that you should compute those things, at least not yet, I wanted you to visualize that. A piece of paper and a pencil would do it.

You have an angle with the normal vector. But that angle, as you have already discovered, does not define a unique vector for reflection; rather it defines a cone.

To select the real reflected ray, you need another angle, in the tangent plane - the angle with the (projection of the) incident ray, not the normal. You cannot find that angle - it is postulated in the law of reflection. Just think about that, and use that piece of paper!

7. Jul 3, 2013

### szynkasz

The normal vector $\vec{n}$ is a bisector of the angle between the incident and reflected ray. If $\vec{v}_i$ and $\vec{v}_r$ are ray's unit vectors then $(\vec{v}_r-\vec{v}_i)$ is parallel to $\vec{n}$:

$\begin{cases}(\vec{v}_r-\vec{v}_i)\times\vec{n}=\vec{0}\\|\vec{v}_r|=1\end{cases}$

8. Jul 4, 2013

### Saitama

$\vec{v_r}=x\hat{i}+y\hat{j}+z\hat{k}$, $\vec{v_i}=-\hat{i}$ and $\displaystyle \vec{n}=\frac{2\hat{i}+2\hat{j}+\hat{k}}{\sqrt{10}}$

$\vec{v_r}-\vec{v_i}=(x+1)\hat{i}+y\hat{j}+z\hat{k}$.
Solving the cross product, I found relations between x, y and z. Using the second relation i.e $|\vec{v_r}|=1$, I do not get the right answer.

9. Jul 4, 2013

### szynkasz

What is the correct answer?

10. Jul 4, 2013

### voko

I do not think so.

11. Jul 4, 2013

### szynkasz

Why not?

12. Jul 4, 2013

### voko

Since you did not specify the orientation of the unit vectors, there are four possible combinations. Surely all of them cannot have the property you stated. And whether any particular combo does is yet to be seen.

13. Jul 4, 2013

### ehild

The vectors vi and vr are unit vectors of the same directions as the wave vectors of the incident and reflected waves, respectively.

ehild

Last edited: Jul 4, 2013
14. Jul 5, 2013

### ehild

Vi has to be a unit vector, why do you multiply it by 2???
Remember the figure demonstrating reflection. n, the normal vector points outward. Your normal points inward, toward the centre of the sphere.
Vi points toward the surface and Vr points away from it. θ is the angle less than 90° enclosed by both -Vi and Vr.
Otherwise your method is all right, but you need to use the condition that both Vi and Vr are unit vectors. And it would be much easier to use the relation Vr-Vi=2cos(θ)n (where n is the normal unit vector).

ehild

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• ###### vivr.JPG
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Last edited: Jul 5, 2013
15. Jul 5, 2013

### Saitama

Okay, I will rewrite everything.
$\vec{v_i}=-\hat{i}$, $\displaystyle \vec{v_r}=\frac{x\hat{i}+y\hat{j}+z\hat{k}}{\sqrt{10}}$. Referring to your sketch,
$$\vec{v_i}\cdot \frac{(2\hat{i}+2\hat{j}+\hat{k})}{3}=\cos \theta$$
$$\Rightarrow \cos \theta=\frac{-2}{3}$$
Dot product of reflected and incident ray:
$$-\hat{i}\cdot(x\hat{i}+y\hat{j}+z\hat{k})=\cos 2\theta$$
$$\Rightarrow x=\frac{1}{9}$$
Dot product of normal and reflected ray:
$$\frac{(2\hat{i}+2\hat{j}+\hat{k})}{3} \cdot (x\hat{i}+y\hat{j}+z\hat{k})=\cos \theta$$
Substituting x and $\cos \theta$,
$$\frac{2}{9}+2y+z=-2$$
$$\Rightarrow 2y+z=-\frac{20}{9}$$
Also,
$$\vec{v_r}-\vec{v_i}=2\cos\theta \vec{n}$$
$$\Rightarrow (x+1)\hat{i}+y\hat{j}+z\hat{k}=\frac{-4}{3}\frac{(2\hat{i}+2\hat{j}+\hat{k})}{3}$$
If I compare both the sides,
$$x+1=-\frac{8}{9}$$
Solving for x doesn't give 1/9.

16. Jul 5, 2013

### ehild

Look at my sketch more carefully. θ<90° is the angle between the incident ray and the normal. $-\vec v_i\cdot\vec n=\cosθ=2/3$. The angle between vi and the normal is 180-θ.

ehild

Last edited: Jul 5, 2013
17. Jul 5, 2013

### voko

There is a simpler approach. Let $Q$ be the point where the ray strikes the surface. Then $\vec{PQ}$ is the incident unit vector, $\vec{QR}$ is the reflected unit vector, $\vec{QS}$ is the unit normal vector. Consider the line passing through $P$ and $R$. This line intersects with the normal at $T$. Then the law of reflection implies that $\vec{PT} = \vec {TR}$, thus $\vec{QR} = \vec{QT} + \vec{TR} = \vec{QT} + \vec{PT} = \vec{QT} + \vec{PQ} + \vec{QT} = \vec{PQ} + 2\vec{QT}$.

18. Jul 5, 2013

### ehild

Let Pranav work on his own solution.

ehild

19. Jul 6, 2013

### Saitama

Thanks ehild!

The correction gives the right result.

Nice one voko but I don't see why I can't get the right answer from this.
$$\vec{QR}=x\hat{i}+y\hat{j}+z\hat{k}$$
$$\vec{PQ}=-\hat{i}$$
$$\vec{QT}=\frac{2\hat{i}+2\hat{j}+\hat{k}}{3}$$
Substituting,
$$(x+1)\hat{i}+y\hat{j}+z\hat{k}=2\left(\frac{2\hat{i}+2\hat{j}+\hat{k}}{3}\right)$$
From the above relation, I don't get the correct values for x,y and z.

20. Jul 6, 2013

### voko

This is not $\vec{QT}$, this is $\vec{QS}$.

21. Jul 6, 2013

### Saitama

Got it, thanks voko!

22. Jul 6, 2013

### ehild

Splendid

Now, you can see an other hint on my sketch. (It is similar to voko's method.) The red vector is -vi+vr=2ncosθ. n=(2i+2j+k)/3, cosθ=2/3, so vr=(xi+yj+zk)=2ncosθ+vi.

ehild

23. Jul 6, 2013

### Saitama

Typo?

Thanks ehild!

24. Jul 6, 2013

### ehild

Why? The vertical components of both vr and -vi are cosθ. They add. The sum points in the direction of the normal, so it is multiplied by the normal unit vector.

whild

25. Jul 6, 2013

### Saitama

Ah sorry, I misread it. The minus sign and v_i are shown in separate lines here so I did not notice it.