Laser Incident to Surface Perpendicularly

In summary: The standing wave does not move in time, only the amplitude of the standing wave changes, while the traveling wave moves to the right.
  • #1
yklin_tux
7
0
Hello I have a question regarding reflected light from a surface of higher index of refraction.

Suppose I have a ideal laser, and an ideal surface with reflection coefficient 0.7 or something like that.

Say I shine my laser directly perpendicular to the surface (at 90 degrees).

I understand some light will go through, but the reflected light will have a 180 degree phase change, and be traveling opposite to the incoming light.

This screams standing wave pattern, but I do not understand the physical interpretation of this.

I get how a standing wave can be formed in a mirror cavity, but in the present situation where I just have light reflecting from a surface, what happens? I know the equation for my electric field will look something like

E = sin(x - wt) + 0.7 sin(x + wt) assuming that k = 1, E0 = 1

And the result of this is a standing wave. In this case, some energy is transmitted into the surface, and the rest is stored in the standing wave?

I understand seeing the reflected light in this case is impossible because it would mean blocking the incoming light, so seeing reflected light is only possible when light is shined at an angle other than 90 degrees.

Actually, plotting the above function in ROOT and making it a histogram, and watching the projections of y (in watching what happens with the wave as time goes on). I see that the wave actually moves, but very slowly? While still modulating amplitude...

I just want to understand what happens in the case of normal incidence because I clearly have two waves traveling in opposite directions with same frequency.
 
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  • #2
Can we obtain an interference pattern in vacuum?

I want to know too..
 
  • #3
yklin_tux said:
E = sin(x - wt) + 0.7 sin(x + wt) assuming that k = 1, E0 = 1
The result is a combination of a traveling wave + a standing wave. To see this, you can rewrite that equation as

E = 0.3sin(x-wt) + 0.7[sin(x - wt) + sin(x + wt)],​

which shows explicitly the traveling wave and standing wave components.

And the result of this is a standing wave. In this case, some energy is transmitted into the surface, and the rest is stored in the standing wave?
Some of the energy is in the wave (standing and traveling components), some is transmitted through the mirror.

Actually, plotting the above function in ROOT and making it a histogram, and watching the projections of y (in watching what happens with the wave as time goes on). I see that the wave actually moves, but very slowly? While still modulating amplitude...
I'm unable to imagine how to visualize the effect myself, but your simulation is probably correct. Perhaps you could try varying the reflectance of 0.7 ... try something really small (0.1?) and something closer to 100%, perhaps 0.9 or 0.95.

asdofindia said:
Can we obtain an interference pattern in vacuum?
Yes. Light travels in a vacuum, so that is no problem.
 
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  • #4
Redbelly98 said:
The result is a combination of a traveling wave + a standing wave. To see this, you can rewrite that equation as

E = 0.3sin(x-wt) + 0.7[sin(x - wt) + sin(x + wt)],​

which shows explicitly the traveling wave and standing wave components.

When you say a traveling and standing wave, what does that mean? That there is some standing wave pattern which is itself moving? I guess this is the part I do not get...
 
  • #5
No, I mean there is a standing wave (which does not move to the left or right), as well as a traveling wave moving to the right. Can you identify which terms in my equation represent the traveling wave, and which represent the standing wave?

E = 0.3sin(x-wt) + 0.7[sin(x - wt) + sin(x + wt)]​

Mind you, I myself am not able to visualize this combined wave, I am just looking at the math terms in the equation.
 
  • #6
E = 0.3sin(x-wt) + 0.7[sin(x - wt) + sin(x + wt)],
= 0.3 sin(x-wt) + 0.7[(sin x cos wt - sin wt cos x) + (sin x cos wt + sin wt cos x)]
= 0.3 sin(x-wt) + 0.7[(sin x cos wt )+ (sin x cos wt )]
= 0.3 sin(x-wt) + 0.7[2(sin x cos wt )]
= 0.3 sin(x-wt) + 1.4 (sin x) (cos wt)
= the traveling wave + the standing wave
 

Related to Laser Incident to Surface Perpendicularly

1. What is a laser incident to surface perpendicularly?

A laser incident to surface perpendicularly refers to a laser beam that is directed at a surface at a 90-degree angle. This means that the laser beam is traveling straight towards the surface and hitting it directly.

2. What are the applications of a laser incident to surface perpendicularly?

A laser incident to surface perpendicularly is commonly used in laser cutting, engraving, and drilling processes. It is also used in scientific research, such as in spectroscopy and microscopy.

3. How does a laser incident to surface perpendicularly affect the surface it hits?

The effects of a laser incident to surface perpendicularly depend on the properties of the surface and the intensity of the laser beam. In general, it can cause heating, melting, or vaporization of the surface material.

4. What safety precautions should be taken when using a laser incident to surface perpendicularly?

It is important to wear appropriate personal protective equipment, such as safety glasses, when working with a laser incident to surface perpendicularly. The laser beam should also be properly contained and directed to avoid accidental exposure to humans or animals.

5. How can the accuracy of a laser incident to surface perpendicularly be ensured?

The accuracy of a laser incident to surface perpendicularly can be ensured by properly calibrating and aligning the laser equipment. The surface should also be clean and free of any obstructions to ensure the laser beam hits the intended target accurately.

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