Red light is incident in air on a 30 o - 60 o - 90 o prism as shown. The inciden

Question

Red light is incident in air on a 30o - 60o - 90 o prism as shown. The incident beam is directed at an angle of 1 = 38.3owith respect to the horizontal and enters the prism at a height h = 19 cm above the base. The beam leaves the prism to the air at a distance d = 52.8 along the base as shown.

1)What is 2, the angle the beam in the prism makes with the horizontal axis?

2)What is n, the index of refraction of the prism for red light?

3)What is 3, the angle the transmitted beam makes with the horizontal axis?4)

4)What is 1,max, the maximum value of 1 for which the incident beam experiences total internal reflection at the horizontal face of the prism?

5)The red beam is now replaced by a violet beam that is incident at the same angle 1 and same height h. The prism has an index of refraction nviolet = 1.41 for violet light. Compare dviolet, the exit distance for violet light, to d, the exit distance for red light.

dviolet < d

dviolet = d

dviolet > d

6)Suppose now that the violet beam is incident at height h, but makes an angle 1,v = 60o with the horizontal. What is 3,v, the angle the transmitted beam makes with the horizontal axis?

Explanation / Answer

1) horizontal distance of left point of prism to incident point,

d' = h/tan30 = 32.91 cm

tan@2 = h / (d-d')

tan@2 = 19 / (52.8 - 32.9)

@2 = 43.7 deg

b)
incident angle, i = 90 - 30 - 38.3 = 21.7 deg

r = 90 - 43.7 - 30 = 16.3 deg

applying snell's law,

ni sini = nr sinr

1 * sin21.7 = nR sin16.3

nR = 1.317

3) 1.317 sin(90- 43.7) = 1 * sinr

r = 72.26 deg

@3 = 90 - r = 17.74 deg

4) for critical angle,

1.317 * sin@c = 1* sin90

@c = 49.4 deg

@2 = 90 - 49.4 = 40.6 deg

r = 90 - 40.6 - 30 = 19.4 deg

1 * sini = 1.317 sin19.4

i = 25.9 deg

@1 = 90 - 25.9 - 30 = 34.05

for maximum value, (@1 will be on other side of normal)

@1_max = (90 - 30) + 34.05 = 94.05 deg

for next parts replace 1.317 wth 1.41

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