# Index/angles of refraction for violot/red light

• frozenguy
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.

## 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?

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?.. 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.

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?

Oops, sorry. Of course θ2=0. So θ1=0.

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)?

Delphi51 said:
Oops, sorry. Of course θ2=0. So θ1=0.

Its cool, ok so $$n_{1}=\frac{n_{0}sin\theta_{0}}{sin\theta_{1}}$$

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 $$\frac{\lambda_{2}}{n_{1}}=\frac{\lambda_{1}}{n_{2}}$$ ?

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?

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.

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..

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.

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..

Can I assume that the index of refraction is 1.0003 in air for red light, violet light, etc?

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.

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.

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 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!

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.

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?!

Ok so θra=0.997deg=phi