How Does Light Travel Through Different Media?

AI Thread Summary
Light travels through different media by refracting at the boundaries, with the direction of travel changing based on the indices of refraction. In this case, a light ray enters water at an angle of 58° from the normal, requiring the application of Snell's Law to determine its path. The correct calculation involves first finding the angle of incidence in water before transitioning to glass. The angle of incidence on the glass is found to be approximately 39.615° after correcting the indices of refraction. Understanding these principles is crucial for accurately determining light's behavior as it moves through various materials.
BuBbLeS01
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Direction of travel in glass...please please HELP!

Homework Statement


A 0.95-cm-thick layer of water stands on a horizontal slab of glass. A light ray in the air is incident on the water 58° from the normal. What is the ray's direction of travel in the glass? (Give the answer in degrees.)


Homework Equations


Nwater = 1.33
Nglass = 1.5

The Attempt at a Solution


I thought this was...
sin theta 1/sin theta 2 = n2/n1
1.33/1.5 * sin 58 = 0.7519 then take the inverse of sin 0.7519 = 48.758 degrees
 
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The ray is incident on the water at 58 degrees, not the glass.

HINT: You know the angle that the ray is incident on the water. You need to use that to find the angle it is hitting the glass at.
 
so that's just 90-58 = 32 right?
 
BuBbLeS01 said:
so that's just 90-58 = 32 right?

Not quite.

The light is refracted when it passes from the air to the water. So, you are going to have to use Snell's Law and then some geometry to find the angle of incidence on the glass.
 
oh so n1*sin 1 = n2 * sin 2
sin^-1 (1.33 * sin 58 / 1.5) = 48.758 degrees
 
You indices of refraction are wrong. Remember that your two medium for this part of the problem are air and then water.
 
woops
sin^-1 (1.00 * sin 58 / 1.33) = 39.615 degrees
 
Looks good. Now you should be able to go about the problem as you did initially, but now you have the right incident angle for the glass.
 
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