College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 22 - Reflection and Refraction of Light - Learning Path Questions and Exercises - Exercises - Page 774: 9


$alpha=90^{\circ}$ for any value of $\theta_{i1}$

Work Step by Step

Let PQRS is parallelogram. So, Angle PQR + Angle QRS = $180^{\circ}$ Therefore, $(\theta_{i1}+\theta_{r1})+(\theta_{i2}+\theta_{r2})=180^{\circ}$ or, $2\theta_{i1}+2\theta_{r2}=180^{\circ}$ $\theta_{i1}+\theta_{r2}=90^{\circ}$ or, $\theta_{i1}+alpha-\theta_{i1}=90^{\circ}$ Therefore, $alpha=90^{\circ}$ Hence, for $alpha=90^{\circ}$ and for any value of $\theta_{i1}$ the ray will reflect in the direction from where came i.e. parallel to the incident ray.
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