The 45-45-90 triangular prism shown in the figure has an index of refraction of
ID: 1686495 • Letter: T
Question
The 45-45-90 triangular prism shown in the figure has an index of refraction of 1.43, and is surrounded by air. A ray of light is incident on the left face at an angle of 55.0°. The point of incidence is high enough that the refracted ray hits the opposite sloping side.Through which face of the prism does the light exit?
The bottom (hypotenuse).
The right side.
The left side.
Calculate the angle at which the ray exits the prism, relative to the normal of that surface. Express your answer as a negative number if the angle is clockwise relative to the normal, and a positive number for a counterclockwise angle.
What is the maximum index of refraction for which the ray exits the prism from the right-hand side? (It may help to remember that sin(90° - ?) = cos ?.)
Explanation / Answer
Given refractive index of air n _1 = 1 refractive index of the prism n_2 = 1.43 incident angle i = 55.0 ^ o The refracted angle is n_1 sin ?_1 = n_2 sin ?_2 sin (55) = 1.43 sin ?_2 ?_2 = 34.94^ o Thus the angle at the left interface in the upper triangle is 90 - 34.94 = 55.06^ oTherefore on the right side the angle in the upper triangle is 180 - 90- 55.06 = 34.94^ o S o the incident angle on the right face is 55.06^ o
BUT the critical angle for this prism is sin c = 1 / n c = sin ^ -1 ( 1 / 1.43 ) = 44.37^ o
So this ray is reflected internally. so the reflected angle is 55.06^ o
So the angle in the right hand triangle is 34.94^ o .this reflected ray strikes the lower surface at an angle with respect to the bottom of 180 - 45- 34.94 = 100.06^ o
This means the incident ray on the bottom surface 100.06 -90 = 10.06^ o
A) So the original ray will exist the bottom of the prism and the angle it makes is
n_1 sin ?_1 = n_2 sin ?_2 sin (?_1) = 1.43 sin (10.06) ?_1 = 14.465^ o Which based on your system is a positive angle.
C)For the ray to exist the right hand side we have from the left hand side
sin(55) = n*sin(?) at the rt hand side the incident angle will be (90 - ?)
n*sin(90 - ?) = 1 but sin(90 - ?) = cos(?) so n*cos(?) = 1
if we square both eqns we get
sin^2(55) = n^2*sin^2(?).....(1)
and 1 = n^2*cos^2(?)-.........(2) Now adding these we get
1+sin^2(55) = n^2*(sin^2(?) + cos^2(?)) = n^2
n = ((1+sin^2(55)) ^ ( 1 /2 ) =(1+ .671)^ (1/2 ) = 1.2926 = 1.32 Given refractive index of air n _1 = 1 refractive index of the prism n_2 = 1.43 incident angle i = 55.0 ^ o The refracted angle is n_1 sin ?_1 = n_2 sin ?_2 sin (55) = 1.43 sin ?_2 ?_2 = 34.94^ o Thus the angle at the left interface in the upper triangle is 90 - 34.94 = 55.06^ o
Therefore on the right side the angle in the upper triangle is 180 - 90- 55.06 = 34.94^ o S o the incident angle on the right face is 55.06^ o
BUT the critical angle for this prism is sin c = 1 / n c = sin ^ -1 ( 1 / 1.43 ) = 44.37^ o
So this ray is reflected internally. so the reflected angle is 55.06^ o
So the angle in the right hand triangle is 34.94^ o .this reflected ray strikes the lower surface at an angle with respect to the bottom of 180 - 45- 34.94 = 100.06^ o
This means the incident ray on the bottom surface 100.06 -90 = 10.06^ o
A) So the original ray will exist the bottom of the prism and the angle it makes is
n_1 sin ?_1 = n_2 sin ?_2 sin (?_1) = 1.43 sin (10.06) ?_1 = 14.465^ o Which based on your system is a positive angle.
C)For the ray to exist the right hand side we have from the left hand side
sin(55) = n*sin(?) at the rt hand side the incident angle will be (90 - ?)
n*sin(90 - ?) = 1 but sin(90 - ?) = cos(?) so n*cos(?) = 1
if we square both eqns we get
sin^2(55) = n^2*sin^2(?).....(1)
and 1 = n^2*cos^2(?)-.........(2) Now adding these we get
1+sin^2(55) = n^2*(sin^2(?) + cos^2(?)) = n^2
n = ((1+sin^2(55)) ^ ( 1 /2 ) =(1+ .671)^ (1/2 ) = 1.2926 = 1.32
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