index of refraction?

Question

2. Given λ = 469.8[nm] and θ = 37.9°, then what is the index of refraction to the nearest hundredth?
(Assume Φ=60°.)

PLEASE HELP AND EXPLAIN STEP BY STEP I HAVE OTHER PROBLEMS LIKE THIS, THANKS

Answer 1

To answer this, I have to make a couple of assumptions on your problem as stated.

To calculate the index of refraction, you need to have both the incident angle and resultant angle. The incident angle is the angle the light strikes the surface. The resultant angle is the angle the light takes inside the optic.

Check out: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/fermat.html#c3

On the above website, they use θ1 and θ2. θ1 is the incident angle, θ2 is the resultant angle.

To find the index of refraction, you take n1/n2 = sin θ2/sin θ1.

Presume:
n1 = 1
θ1 = Φ=60°
θ2 = 37.9°

Now, solve for n2.

n2 = n1(sin θ1/sin θ2).
= 1 (sin 60 / sin 37.9)
= .866 / .614
= 1.41

Hope this helps. It's been while since I've had to do this, but I think I got it right...