(a) Explanation Given Info: The radius of Earth is 6378 km and the atmosphere has uniform density and its uniform index of refraction is n and the height is h above the Earth’s surface. The below figure shows the position of Sun at sunrise or sunset Figure (I) The apparent position of Sun is S’ which is formed due to the atmospheric refraction. Rays that comes from the Sun strikes at the point B and bends towards the normal and finally reaches the observer which is at point A. Thus the light seems to be come from S’. From Snell’s law, sin θ a sin θ b = n θ a is angle of incidence. θ b angle of refraction. n is refractive index. Rearrange for sin θ a . sin θ a = n ( sin θ b ) (I) From Figure (I), δ = θ a − θ b (II) δ is angle of deviation. From triangle OAB in Figure (I), sin θ b = R R + n R is radius of Earth (b) To determine Angle of deviation with n = 1.0003 and h = 20 km and compare this to the angular radius of Sun.

University Physics with Modern Physics (14th Edition)

14th Edition

ISBN: 9780321973610

Author: Hugh D. Young, Roger A. Freedman

Publisher: PEARSON

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Dispersion of White Light

Snell’s law is a formula that gives a relation between the angle of incidence and angle of refraction and is given as:

Question

Chapter 33, Problem 33.51P

(a)

To determine

The validation of the relation for angle of deviation is δ=arcsin(nRR+h)−arcsin(RR+h)

(b)

To determine

Angle of deviation with n=1.0003 and h=20 km and compare this to the angular radius of Sun.

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Chapter 33, Problem 33.50P

Chapter 33, Problem 33.52P

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Chapter 33 Solutions

University Physics with Modern Physics (14th Edition)

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The validation of the relation for angle of deviation is δ = arcsin ( n R R + h ) − arcsin ( R R + h )

Solution Summary: The author explains that the relation for angle of deviation is delta =arcsin(

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The validation of the relation for angle of deviation is δ=arcsin(nRR+h)−arcsin(RR+h)

Angle of deviation with n=1.0003 and h=20 km and compare this to the angular radius of Sun.

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Chapter 33 Solutions