Thread: Angle of refraction

Hi, I am having some trouble with this question.

The index of refraction of light passing from air to water is 1.33. If the angle of incidence is 38 deg, find the angle of refraction.

Thank you!

using snell's law we get:

theta = arcsin[(n1/n2)*sin(38)] ~ 55.2

^ I do not quite get Snell's law =( Can you possibly explain a bit of the steps to me.

What I know is that the formula is like this:

sin theta 1 / sin theta 2 = v1 / v2

v1 / v2 = index of refraction

we are given the angle of incidence which the value of v1 .

so it would be

sin 38 deg / sin theta 2 = 1.33 ?_?

I have always thought snells law was:
n1 sin theta1=n2 sin theta2
where n1 is the refractive index of the first substance,
theta1 is the angle of incidence,
n2 is the refractive index of the second substance and
theta2 is the angle of refraction

Therefore, n1=1.00 (absolute refractive index of air)
n2=1.33 (absolute refractive index of water)
theta1=38 deg

so n1 sin theta 1=n2 sin theta 2
1 sin 38=1.33sin theta2
theta2=sin^-1 (sin38/1.33)
theta2=27.57deg

Hope that helps

Originally Posted by trent19

I have always thought snells law was:
n1 sin theta1=n2 sin theta2
where n1 is the refractive index of the first substance,
theta1 is the angle of incidence,
n2 is the refractive index of the second substance and
theta2 is the angle of refraction

Therefore, n1=1.00 (absolute refractive index of air)
n2=1.33 (absolute refractive index of water)
theta1=38 deg

so n1 sin theta 1=n2 sin theta 2
1 sin 38=1.33sin theta2
theta2=sin^-1 (sin38/1.33)
theta2=27.57deg

Hope that helps

I second that.
I've always thought snell's law was n1 sin(theta)1 = n2 sin(theta)2 as well!
So the question should provide you with index of refraction for both substances.