# Free end reflection question

1. May 12, 2015

### roineust

Hello there,
Here is an elementary explanation about 'Free end reflection':

http://www.physicsclassroom.com/mmedia/waves/free.cfm

My question is:
Say we radiate from a certain direction a rigid not hollow object, that is surrounded by gas or by vacuum, with electromagnetic or other type of radiation that can pass through the object partially or more than that .
If we check the internal reflection direction of that radiation residues inside the object - will we find out that that radiation reflects inside the object at the same direction it came from or inverted?

Thanks.

Last edited: May 12, 2015
2. May 13, 2015

### Staff: Mentor

I don't understand the question. Can you draw a sketch?

There will be light (or whatever) reflected back at the "exit" (back in the direction of the source), a part of that will get reflected in the opposite direction again and so on, but with a probability that goes to zero for many reflections.

3. May 13, 2015

### roineust

Here is the sketch. No kind of radiation reflects back inside a radiated object? or hardly at all?

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• ###### Object inner reflection.jpg
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4. May 14, 2015

### Staff: Mentor

The lower right "?" ray will be present in general, together with the left "?" ray. The upper "?" ray is incoming light only.

5. May 14, 2015

### roineust

Why is it that D will be present in general and not C?
Isn't it considered a 'Free end reflection' case?

#### Attached Files:

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Last edited: May 14, 2015
6. May 14, 2015

### Staff: Mentor

You don't have a one-dimensional problem here. Incoming angle = deflection angle applies for two-dimensional problems.

7. May 14, 2015

### roineust

Do you mean that the question regarding the BEAM angle is a one dimentional question and that the answer is D angle de/reflection of the ray and not a C angle.

And that the question regarding the 'Free end reflection' is a different kind of question - a two dimentional question and has to do only with the blue or red kind of ray WAVE (phase) and that the answer to this other question, is that because of 'Free end reflection' it would be the red wave phase and not the blue wave phase which reflects?

#### Attached Files:

• ###### Object inner reflection 3.jpg
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Last edited: May 14, 2015
8. May 14, 2015

### Drakkith

Staff Emeritus
By inverted, are you talking about the phase of the reflected EM wave?

9. May 14, 2015

### roineust

Yes, the phase of the reflected EM wave.

10. May 14, 2015

### Staff: Mentor

No. The idea "the reflection is opposite to the incoming light" comes from a one-dimensional case, where there are just two directions. That idea does not work for two-dimensional cases like the one in the sketch.

11. May 14, 2015

### roineust

I did not understand - can you please refer me to a really plain explanation of the two dimentional case vs. the one dimentional case?

12. May 14, 2015

### Staff: Mentor

Forget the one-dimensional case, that is my point all the time. Forget the "free end".

Light is reflected in the same way a pool ball bounces from a border, for example. Always the "D" path.

13. May 23, 2015

### roineust

Just to be clear about that - even if it is not a pool, in the sense that there are water surrounded by a denser material - But a pool filled with a denser material, the type that usually the edges are made of (e.g. concrete) and a less denser material (e.g. water, gas or vacuum) is what the edges are made of and the radiation travels within the denser material in the middle of the 'pool', until it reaches the less denser edge? Still only the 'D' path?

Last edited: May 23, 2015
14. May 24, 2015

### theodoros.mihos

You can think about wave equation solution for infinity wave. On different materials the wave have different wave vector k, then the continuity of k components to all axis give the reflection - refraction Snell's law.

15. May 24, 2015

### Staff: Mentor

Plus B, yes. No C. And the D light will have another partial reflection at the lower edge, of course.

Light does not travel well within concrete.