Snell's Law & Light help!

  • Thread starter hshphyss
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  • #1
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Can anyone help me with these questions? They are due @ 10 p.m. tonight, thank-you.

1) A ray of light enters the top of a glass of water at an angle of 37° with the vertical. What is the angle between the refracted ray and the vertical?

At first I thought I could just do 90-37, but that did not work, and then I tried solving for angles of incidence and refraction but they both did not work. Snells Law=ni sin theta i=nr sin theta r


15-20.gif

2)Light strikes the surface of a prism, n = 1.4, as shown. If the prism is surrounded by a fluid, what is the maximum index of refraction of the fluid that will still cause total internal reflection within the prism?

i'm not sure what to do for this problem, does snell's law still apply, would i solve for the critical angle?
 

Answers and Replies

  • #2
hshphyss said:
Can anyone help me with these questions? They are due @ 10 p.m. tonight, thank-you.
1) A ray of light enters the top of a glass of water at an angle of 37° with the vertical. What is the angle between the refracted ray and the vertical?
At first I thought I could just do 90-37, but that did not work, and then I tried solving for angles of incidence and refraction but they both did not work. Snells Law=ni sin theta i=nr sin theta r
Snell's law is definitely the way to go on this one. What went wrong when you tried to use it? Snell's law can be written as:
[tex]\frac{n_2}{n_1}=\frac{\sin\theta_1}{\sin\theta_2}[/tex]
Theta 1 is the angle the ray entering makes with the vertical and is given as 37°. n1 is probably that of air which you can take to be 1. n2 is the index of refraction of water, which you can look up. Theta2 is what you want.

hshphyss said:
2)Light strikes the surface of a prism, n = 1.4, as shown. If the prism is surrounded by a fluid, what is the maximum index of refraction of the fluid that will still cause total internal reflection within the prism?
i'm not sure what to do for this problem, does snell's law still apply, would i solve for the critical angle?
Snell's law tells us, after solving for theta 2, that:
[tex]\theta_2= \sin^{-1}{(\frac{n_1}{n_2}\sin\theta_1)}[/tex]
Look inside the inverse sin. You know that you can't take the inverse sin of anything geater than 1 or less than -1. You know theta1 is 45°. Now you can solve for the value of n2 that makes whats inside the inverse sin go to 1. You know this is the maximum value that will cause total internal reflection because if n2 were any greater that would make what's in the inverse sin less than 1, which would mean that refraction is allowed.
 
  • #3
lightgrav
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That is, you USE Snell's Law to solve for the critical angle!

In optics, angles are ALWAYS measured from the Normal
(perpendicular, piercing the surface)

(Michael, got a little carried away, didn't you ...?)
 
  • #4
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so where would 1.4 be incorporated? theta2=sin^-1(1.4/n2 sin45) ??i have two unknowns what would theta 2 be?? thank-you
 
  • #5
lightgrav
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using the form : [tex]n_1 sin (\theta_1) = n_2 sin (\theta_2) [/tex] ,
the "1"s all refer to one side of the boundary, the "2"s all refer to the other side.
For the light to just totally reflect, the angle in the fluid outside the glass
has to be = 90 degrees ... (that is, have the largest sin possible).
 
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  • #6
andrevdh
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Total internal reflection occurs when a beam moves from a less dense (smaller n) to a more dense (larger n) medium. In such a case the refraction angle is larger than the incident angle. If the incident angle is therefore increased the refraction angle will eventually become 90 degrees, meaning that the refracted beam is moving along the interface of the two media. If the incident angle is increased beyond this critical incident angle the refracted beam dissappears an we have only reflected light. Applying Snell's law for this limiting case we therefore come to
[tex]n_g\sin(45^o)=n_f\sin(90^o)[/tex]
from which the index of refraction for the fluid can be determined. If the index of refraction of the fluid is reduced below this value the refraction angle will become less than 90 degrees - see my other recent post on index of refraction - resulting in a refracted beam with no subsequent total internal reflection. Why do we want total internal reflection? No light is lost to refraction, all of the light is concentrated in the image transferrence by the prism.
 
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