Writing Quiz 7 question 2 (three points) The figure below is similar to that sho
ID: 1566680 • Letter: W
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Writing Quiz 7 question 2 (three points) The figure below is similar to that shown in problem 3 in intermet homework assignment HW 8b. However, the questions asked below differ from those ofproblem 3. In this figure, a light ray is incident upon the plastic from air, which hasindex of refraction equal to one. The plastic has an index ofrefraction cqual to 135. (a) For the given indices of refraction, what value ofangle 6A will result in an angle 90°? (b) Identify what range of values for the angle will cause total intermal reflection at the boundary of the second surface of intersection. Write your answer in the form lowest angle value 26 2 largest angle value Explain how you determined this range of values. (c) Carefully explain why no value of angle will cause total intenmalreflection at the boundary of the first surface of intersection.Explanation / Answer
a) referactive index = 1.35
for theta_2 to be 90 degree, the angle of incidence on the second surface must be such that it satisfies:
1.35 sin(i) = 1*sin(90)
i = 47.79 degree
the angle of refraction at first surface = 90-47.79 deg = 42.21 deg
1 sin(theta_1) = 1.35*sin(42.21 deg)
theta_1 = 65.09 degree
b)
For this, the angle of incidence on the second surface i should be:
47.79 deg < i <90 deg
The angle of refraction at first surface, r should be .
47.79 deg < 90-r <90 deg
-42.21 deg < -r < 0 deg
0<r<42.21 deg
taking sine both side,
0 < sin(r) < sin(42.21 deg)
or 0 < 1.35 sin(r) < 1.35 sin(42.21 deg)
0 < sin(theta_1) <0.90699733
0<theta_1<65.094 degree
c)
Since the ray is going from rare to dense medium hence the TIR won't happen in this case.
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