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I think there may be a difficult experiment that I don't know, I want to ask it.

We know if the angle of incidence is greater than 0 degree , we can actual see the light beams reflected in the same degree. However, if the angle of incidence is 0 degree, they said it will reflect at 0 degree also. I'd like to know what experiment did they carry out? :confused:

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How can scientist know that when a line pass along the normal, from less denser to denser , it still reflects along the normal [/b'] ?

 

From what I was told in physics class, this is true. When the angle is not 0o with respect to the normal, the tip of the light enters the new material at different times. The light can be visualized as a long rectangle. When one corner of this rectangle hits the new material (becuase it is entering at an angle) it changes direction. The front of the light beam continues to enter until the other corner reaches the new material and it has all bent. The light continues for however long the beam is until the two corners at the other end enter and the light has bent completely. When the beam is moving along the normal, both corners of the beam enter at the same time and so they aren't bent. The beam moves through until the other end passes completely into the new material.

 

I don't know how much sense that made. I don't even trust it, I'm just reiterating what I was told. I never pictured a beam of light as having mass or width and so I don't know how it can be pictured as a rectangle. This is just what I know off the top of my head.

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I think there may be a difficult experiment that I don't know' date=' I want to ask it.

We know if the angle of incidence is greater than 0 degree , we can actual see the light beams reflected in the same degree. However, if the angle of incidence is 0 degree, they said it will reflect at 0 degree also. I'd like to know what experiment did they carry out? :confused:[/quote']

 

Lasers wouldn't work as advertised if that wasn't true.

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This is a very simple experiment to perform:

 

You have a polarized source incident on a mirror along the normal. Between the source and the mirror you have placed a polarizing beam splitter (PBS) in such a way that it transmits the light from your source. Now place a half waveplate between the PBS and mirror in such an orientation that it rotates the lights polarization 45 degrees when passing through it before it hits the mirror. On the way back from the mirror (assuming light is reflected back along the normal as you were questioning) the polarization of the light is rotated an other 45 degrees. When the light reaches the PBS the polarization is normal to the incident ligth, meaning the light will reflect at the PBS instead of transmitt. Place a detector at the reflecting output off the PBS, and hey presto, you will find that you measure the same intensity as the output of your source, showing that light, indeed, reflects back along the normal. (This assumes all components do not scatter or absorb any of the light incident on them)

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This is a very simple experiment to perform:

 

You have a polarized source incident on a mirror along the normal. Between the source and the mirror you have placed a polarizing beam splitter (PBS) in such a way that it transmits the light from your source. Now place a half waveplate between the PBS and mirror in such an orientation that it rotates the lights polarization 45 degrees when passing through it before it hits the mirror. On the way back from the mirror (assuming light is reflected back along the normal as you were questioning) the polarization of the light is rotated an other 45 degrees. When the light reaches the PBS the polarization is normal to the incident ligth' date=' meaning the light will reflect at the PBS instead of transmitt. Place a detector at the reflecting output off the PBS, and hey presto, you will find that you measure the same intensity as the output of your source, showing that light, indeed, reflects back along the normal. (This assumes all components do not scatter or absorb any of the light incident on them)[/quote']

 

Would you draw a diagram?

Too difficult..

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