A ray of light in air is incident on the mid-point of a heavy flint glass prism surface at an angle of 20º with the normal. for
the flint glass prism, n = 1.60, and the prism apex angle is 35º. what angle does the ray make with respect to the normal as it exits the heavy flint glass prism on the right (assume the prism is surrounded by air)?
Assuming that you have a triangular prism, the ray of light will undergo refraction twice. The first time is the transition from air to flint glass on the entry face, and the second time is the transition from the flint glass to air from the exit face. With the available data, there are two possible solution since saying "20Âş from the normal" isn't enough information. Depending upon which side of the normal that 20 degrees is, the interior triangle will have the angles of 35, 90-r, and 55+r, or 35, 90+r, 55-r degrees where r is the angle from the normal after the 1st refraction. I will provide both possible solutions and you'll need to actually select the correct one based upon the actual geometry which I don't know because you didn't provide the figure or diagram that you were provided with.
The equation for refraction is:
(sin a1)/(sin a2) = n1/n2
where
a1,a2 = angles from the normal to the surface.
n1,n2 = index of refraction for the transmission mediums.
For this problem, we've been given an a1 of 20Âş and an n1 of 1.60. For n2, we will use air which at STP has an index of refraction of 1.00029. So
(sin a1)/(sin a2) = n1/n2
(sin 20)/(sin a2) = 1.00029/1.60
0.342020143/(sin a2) = 0.62518125
0.342020143 = 0.62518125(sin a2)
0.547073578 = sin a2
asin(0.547073578) = a2
33.16647891 = a2
So the angle from the normal INSIDE the prism is 33.2Âş. The resulting angle from the surface of the entry face will be either 90-33.2 or 90+33.2 depending upon the geometry. So the 2 possible triangles will be either 35Âş, 56.8Âş, 88.2Âş or 35Âş, 123.2Âş, 21.8Âş. with a resulting angle from the normal of either 1.8Âş or 68.2Âş. I can't tell you which one is correct since you didn't tell me which side of the normal the incoming ray came from. So let's calculate both possible exits.
1.8Âş
(sin a1)/(sin a2) = n1/n2
(sin 1.8)/(sin a2) = 1.6/1.00029
0.031410759/(sin a2) = 1.599536135
0.031410759= 1.599536135(sin a2)
0.019637418= sin(a2)
asin(0.019637418) = a2
1.125213477 = a2
68.2Âş
(sin a1)/(sin a2) = n1/n2
(sin 68.2)/(sin a2) = 1.6/1.00029
0.928485827/(sin a2) = 1.599536135
0.928485827 = 1.599536135(sin a2)
0.58047193 = sin a2
asin(0.58047193) = a2
35.48374252 = a2
So if the interior triangle is acute, the answer is 1.13Âş and if the interior triangle is obtuse, the answer is 35.48Âş
