Index/angles of refraction for violot/red light

In summary: Red=0.993degSo that's it?..In summary, white light is incident onto a 30.0[deg] prism at the 40.0[deg] angle shown in the figure. Violet light emerges perpendicular to the rear face of the prism. The index of refraction of violet light in this glass is 2.00% larger than the index of refraction of red light. By using the law of refraction, it was determined that the angle at which red light emerges from the rear face is approximately 0.993 degrees.
  • #1
frozenguy
192
0

Homework Statement


White light is incident onto a 30.0[deg] prism at the 40.0[deg] angle shown in the figure. Violet light emerges perpendicular to the rear face of the prism. The index of refraction of violet light in this glass is 2.00% larger than the index of refraction of red light. At what angle (phi) does red light emerge from the rear face?

Homework Equations


on paper


The Attempt at a Solution


So This is what I was thinking, but not sure if I can use those equations like I did.

Also, if the violet light is refracting perpendicular, that means it was not perpendicular to the right side before it exited/refracted, right?..
Thanks for your help!
231hw_1.jpg
 
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  • #2
it was not perpendicular to the right side before it exited/refracted, right?
You need the refraction formula with the angles in it.
n1*sin(θ1) = n2*sin(θ2)
In the special case where θ2 = 90 degrees, it doesn't matter what the indices are; you can immediately calculate θ1.

Having that θ1 for incidence on the right side, you can work your way around the triangle to get the angle of refraction into the left side, then use the formula again to find the index of refraction for violet light in the glass.
 
  • #3
Delphi51 said:
You need the refraction formula with the angles in it.
n1*sin(θ1) = n2*sin(θ2)
In the special case where θ2 = 90 degrees, it doesn't matter what the indices are; you can immediately calculate θ1.

Having that θ1 for incidence on the right side, you can work your way around the triangle to get the angle of refraction into the left side, then use the formula again to find the index of refraction for violet light in the glass.

Hey Delphi51 thanks for responding! I really want to get this problem.

So how is θ2=90[deg] (if θ2 is the angle between the violet ray and the normal to the right surface)? I assumed it was 0[deg] to the normal.

And if θ2 was 90deg, sin(θ1)=n2/n1 right? How do the indices not matter?
 
  • #4
Oops, sorry. Of course θ2=0. So θ1=0.
 
  • #5
What is the angle of incidence for the left-hand surface?

What is the angle of refraction for the left-hand surface (for violet light)?
 
  • #6
Delphi51 said:
Oops, sorry. Of course θ2=0. So θ1=0.

Its cool, ok so [tex]n_{1}=\frac{n_{0}sin\theta_{0}}{sin\theta_{1}}[/tex]

So n1 is the index of refraction for the prism right?
Is there a standard wavelength for violet light that I'm supposed to use (400nm?) to find the index of refraction? Can I use [tex]\frac{\lambda_{2}}{n_{1}}=\frac{\lambda_{1}}{n_{2}}[/tex] ?
 
  • #7
SammyS thanks for responding!
SammyS said:
What is the angle of incidence for the left-hand surface?
I got 50[deg] for that

SammyS said:
What is the angle of refraction for the left-hand surface (for violet light)?

I got 30[deg] for that..

When does the light turn into colors? during refraction back into the air or during refraction into the prism?
 
  • #8
Ok so I've been doing a little research. Looks like the light disperses inside the prism with their respective angles of refraction, and then again when leaving the prism at another angle of refraction.
 
  • #9
Oh. Ok.. So since the violet light is parallel with the base of the triangle, I'm finding nviolet with n0sinθ0=n1sinθ1 on the left side considering θ1 to be 30deg from normal of left surface and θ0 to be 50deg.

The question states nviolet = 1.02nred right? I'm going to work with that..
 
  • #10
Did you get the number for the violet index of refraction (from knowing the angles of incidence and refraction for the first surface)? Then you can get the red index and do the refraction at the second surface. You don't need to work with wavelengths at all.
 
  • #11
Delphi51 said:
Did you get the number for the violet index of refraction (from knowing the angles of incidence and refraction for the first surface)? Then you can get the red index and do the refraction at the second surface. You don't need to work with wavelengths at all.

I get 50 deg for my red angle to the norm of the right surface.. Sound right? I'll upload my work..
 
  • #12
Can I assume that the index of refraction is 1.0003 in air for red light, violet light, etc?
 
  • #13
Entering the glass on the left side:
n1*sin(θ1) = n2*sin(θ2)
1*sin(50) = n2*sin(30)
Solve for n2. This is the index of refraction of the prism for violet light.
Then use your nviolet = 1.02*nred to find the index for red light.
 
  • #14
Delphi51 said:
Entering the glass on the left side:
n1*sin(θ1) = n2*sin(θ2)
1*sin(50) = n2*sin(30)
Solve for n2. This is the index of refraction of the prism for violet light.
Then use your nviolet = 1.02*nred to find the index for red light.

Yeah I did that.. Lemme upload it.
 
  • #15
here we go..
nvg is nviolet,glass
nrg is nred,glass
the ra is red,air etc.

Goal is to find phi which in this case I labeled θra

231hw_1E.jpg
 
  • #16
frozenguy said:
Can I assume that the index of refraction is 1.0003 in air for red light, violet light, etc?


I must assume this right? Otherwise the colors would go their own separate ways!
 
  • #17
Okay on the 30.7, but not the last step.
You must begin with 30.7 and work your way around the triangle to find the angle of incidence on the right hand surface. It should be very close to zero. And after you run it through the law of refraction, very close to zero coming out of the prism.
 
  • #18
Delphi51 said:
Okay on the 30.7, but not the last step.
You must begin with 30.7 and work your way around the triangle to find the angle of incidence on the right hand surface. It should be very close to zero. And after you run it through the law of refraction, very close to zero coming out of the prism.

Ok I get that.. 90-30.7=59.3

And then that big angle is 180-60=120deg
So 180-59.3-120= 0.7 deg incident to right side?!
 
  • #19
Ok so θra=0.997deg=phi
 

What is the index of refraction for violet light?

The index of refraction for violet light is approximately 1.53.

What is the index of refraction for red light?

The index of refraction for red light is approximately 1.52.

How does the angle of incidence affect the angle of refraction for violet light?

The angle of incidence and the angle of refraction for violet light are related by Snell's Law: n1sinθ1 = n2sinθ2. This means that as the angle of incidence increases, the angle of refraction also increases.

What material has the highest index of refraction for red light?

Diamond has the highest index of refraction for red light, with a value of approximately 2.42.

What is the relationship between the index of refraction and the speed of light?

The index of refraction is inversely proportional to the speed of light in a given medium. This means that as the index of refraction increases, the speed of light decreases.

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