Space, index of refraction n =1 Atmosphere, index of refraction n = 1.0003 Not to Scale! tº theta 0.5° Sun The top edge of the sun is an angle theta-88.8 degrees from the vertical as seen from space. The angular size of the sun is 0.5 degrees. When the rays from the sun enter our atmosphere (assumed to be a sharp boundary for this problem), they refract. The index of refraction of air can be taken to be 1.0003 (uniform with altitude for simplicity). What is the new angular size of the sun, as seen from inside the atmosphere? degrees

Space, index of refraction n =1 Atmosphere, index of refraction n = 1.0003 Not to Scale! tº theta 0.5° Sun The top edge of the sun is an angle theta-88.8 degrees from the vertical as seen from space. The angular size of the sun is 0.5 degrees. When the rays from the sun enter our atmosphere (assumed to be a sharp boundary for this problem), they refract. The index of refraction of air can be taken to be 1.0003 (uniform with altitude for simplicity). What is the new angular size of the sun, as seen from inside the atmosphere? degrees

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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QUESTION 21 Calculate the critical angle for the light traveling from glass (nglass=1.48) to air (nair=1.00). Express your answer in degrees. Normal format with 3 SF.
A1. As shown in Figure A1, a light ray is incident normal to one face of a 30°-60°-90° prism (n = 1.50) that is immersed in water (n = 1.33). Determine the exit angle 03. (a) Determine the exit angle 03. (b) Determine the critical angle for the prism-water system. n2 60.0° n1 Figure Al 021 30.0% D 103
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2. A ray of light in glass (n₁=1.5) enters another material (n2) at a 70° angle; and emerges into the new material at a 45° angle. Find the index n2. n₁=1.5 70° N2 45°
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