Determining Beam Diameter in Glass Using Index of Refraction

AI Thread Summary
The discussion revolves around calculating the diameter of a laser beam as it transitions from air into glass with a specific index of refraction. Participants debate whether the beam's diameter remains constant or changes due to the angle of incidence and refraction. It is suggested that the diameter may appear larger due to the angle at which the beam strikes the glass, effectively increasing the illuminated area. The importance of accurately interpreting the angles and understanding the geometry involved is emphasized, with some concluding that the index of refraction may not significantly impact the diameter in this scenario. Overall, the conversation highlights the complexities of beam behavior at media boundaries and the need for precise geometric considerations.
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Homework Statement


A non-expanding/non-contracting laser beam having a diameter 0.132 m in air strikes a piece of glass (ng = 1.62) at an angle of 52°. What is the diameter of the beam, in m, in the glass?


Homework Equations


snell's law, n1sin(θ1)=n2sin(θ2)


The Attempt at a Solution


I get the feeling that the diameter should remain the same. I don't understand why it would be any different than that. Please let me know if I am wrong
 
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heycoa said:

Homework Statement


A non-expanding/non-contracting laser beam having a diameter 0.132 m in air strikes a piece of glass (ng = 1.62) at an angle of 52°. What is the diameter of the beam, in m, in the glass?

Homework Equations


snell's law, n1sin(θ1)=n2sin(θ2)

The Attempt at a Solution


I get the feeling that the diameter should remain the same. I don't understand why it would be any different than that. Please let me know if I am wrong

"I get a feeling" is not a prof. Prove that it is the same. A drawing can help.

ehild
 
Well I did do a drawing. And what I did was picture the "walls" of a cylinder of laser light as independent rays. Those transmitting through the surface of glass should remain parallel due to Snell's law. But I am not sure if I am allowed to do this.
 
The rays stay parallel with each other, but make an angle with the incident beam. How is the diameter of the beam defined? ehild
 
I copied the question exactly, so that info in the op is all I got. In my drawing, the beam itself is 38 (90-52) degrees from the normal vector from the glass. Since the beam in my picture is a set of two parallel lines, then each of these lines have the same incident angle, which to me would mean that they transmit through the glass at exactly the same angle, leaving the beams parallel and thus, the diameter constant throughout the whole problem.

I feel my logic is sound however, I can't help but wonder if there is some principle about a non converging, non diverging laser traveling from media. I mean, is it at all possible that the laser converges or diverges after it travels through the glass? It seems very possible that it does, I am just missing the "why" and "how".
 
Have you tried drawing an exaggerated picture where the incoming beam is about 70 degrees to the normal while the refracted beam is about 20 degrees to the normal?

Following echild, make sure you are interpreting the diameters correctly.
 
Oh i think i understand now. Its because the beam is hitting at an angle, so the diameter of the transmitted beam will basically equal the surface area of the glass that is being essentially lit by the laser, which will be larger than the diameter.
 
Is it of your guys' opinion that the 52 degrees (that the problem stated) is an incident angle or the angle with respect to the glass surface?
 
Could anyone please check my work in the picture and tell me if you think I am right? For the new diameter y i got 0.168 m. It seems that the index of refraction of the glass is completely irrelevant and that this is just a simple geometry problem. Please let me know what you think

https://docs.google.com/drawings/d/1gSaU2T_v0thhg_43DWlEjgoSG0Wvor_kzg2gemF5-0A/pub?w=960&h=720
 
  • #10
Note that y is not the diameter of either beam. Can you draw the diameter of the refracted beam similarly to the way you drew the diameter of the incident beam?
 
  • #11
shouldnt y be correct though? I don't understand how the refracted beam could be different than that what I got.
 
  • #12
Note that you drew the diameter of the incoming beam as a line that is perpendicular to the rays of light. Shouldn't you do that also for the refracted beam?
 
  • #13
You are absolutely right, Thank you very much
 
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