Snell's law of refraction problem

[rockind78]

Hello everyone,

I am reviewing for a test tomorrow, and I seem to have come across what would seem like an easy problem, but its answer has eluded me for the time being. I am hoping you guys can help.

Horizontal rays of red light (660nm in a vacuum) and violet light (410nm in a vacuum) are incident on the flint-glass prism showin in the drawing. The indices of refraction fro the red and violet light are 1.662 and 1.698 respectively. What is the angle of refraction for each ray as it emerges from the prism?

The answers are 32.8* for the red and 33.6 for the violet. I just don't know how to arrive (correctly) at these answers.

*Image shows a right triangle, with angles 19* and 71*. The right angle and the 71* angle compose the base.

Any help would be greatly appreciated. Thank you.

 

[Doc Al]

Originally posted by rockind78
Horizontal rays of red light (660nm in a vacuum) and violet light (410nm in a vacuum) are incident on the flint-glass prism showin in the drawing. The indices of refraction fro the red and violet light are 1.662 and 1.698 respectively. What is the angle of refraction for each ray as it emerges from the prism?

Review Snell's law of refraction:
n₁sinθ₁ = n₂sinθ₂

 

[M98Ranger]

Hey there,

Snell's law of refraction can definitely be tricky, but let's break it down step by step to solve this problem. First, let's define some variables:

n1 = index of refraction for the incident medium (in this case, vacuum)
n2 = index of refraction for the refracted medium (in this case, flint-glass)
θ1 = angle of incidence (the angle between the incident ray and the normal)
θ2 = angle of refraction (the angle between the refracted ray and the normal)

Now, let's use Snell's law: n1sinθ1 = n2sinθ2

For the red light: n1 = 1 (since it is in a vacuum) and n2 = 1.662, so we have: sinθ1 = 1.662sinθ2

Similarly, for the violet light: n1 = 1 and n2 = 1.698, so we have: sinθ1 = 1.698sinθ2

Now, we know that the angle of incidence for both the red and violet rays is 71 degrees (since it is the complementary angle to the 19 degree angle given in the drawing). So, we can plug in this value for θ1 in both equations and solve for θ2.

For the red light: sin(71) = 1.662sinθ2
θ2 = sin^-1(0.674) = 42.4 degrees

For the violet light: sin(71) = 1.698sinθ2
θ2 = sin^-1(0.666) = 40.3 degrees

However, these angles are the angles of refraction inside the prism. To find the angle of refraction as it emerges from the prism, we need to use Snell's law again, but this time with n1 = 1.662 and n2 = 1 (since the light is now going from the flint-glass back into vacuum). This gives us:

sinθ2 = 1sinθ1 = sin(42.4) = 0.682
θ2 = sin^-1(0.682) = 41.3 degrees

Similarly, for the violet light, we have:
sinθ2 = 1sinθ1 = sin(40.3) = 0.656
θ2 =