An optical fiber with an index of refraction of 1.5 is used to transport light of wavelength 400 nm.What is the critical angle for light to transport through this fiber without loss if the fiber is immersed in water?

Question

Quizplus · Accepted Answer

The critical angle can be found using Snell's law, $n_1 \sin(\theta_c) = n_2 \sin(90^\circ)$, where $n_1$ is the index of refraction of the optical fiber (1.5), $n_2$ is the index of refraction of water (approximately 1.33), and $\theta_c$ is the critical angle. Solving for $\theta_c$ gives a value close to 60 degrees.

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