Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 23 - Light: Geometric Optics - General Problems - Page 676: 72

Answer

$54.6^{\circ}$.

Work Step by Step

When light is incident at the critical angle, the refracted angle is 90 degrees. Start with the situation of refraction from plastic to air, using equation 23–6. $$sin\theta_C=\frac{n_{air}}{n_{plastic}}$$ Find the plastic’s index of refraction from the information given. $$ n_{plastic}=\frac{n_{air}}{sin\theta_C}=\frac{1.00}{ sin37.8^{\circ}}=1.6316$$ Now find the critical angle when the plastic is immersed in water. $$sin\theta_C=\frac{n_{water}}{n_{plastic}}$$ $$ \theta_C=sin^{-1}\frac{n_{water}}{n_{plastic}}$$ $$ \theta_C=sin^{-1}\frac{1.33}{1.6316}=54.6^{\circ}$$
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