Total internal refraction inside an optical fiber

[HunterDX77M]

Homework Statement

In the attachment below, a ray of light undergoes total internal reflection in an optical fiber such that the angle θ between the ray and the interface is 47°. Assuming the fiber is surrounded by air, what is the minimum index of refraction necessary for this to occur?

Homework Equations

Snell's Law:
$$n_1 sin(\theta_1) = n_2 sin(\theta_2)$$

Assume the index of refraction for air is n = 1.

The Attempt at a Solution

Total internal refraction occurs when the light ray is incident to the air at an angle greater than the critical angle, θ_c. At the critical angle, the refracted light makes a right angle. So . . .

$$n_1 sin(\theta_c) = n_2 sin(90 \circ) = n_2$$
$$n_1 = n_2 \div sin(\theta_c) = 1 \div sin(47 \circ) = 1.37$$

But the correct answer is supposed to be 1.47. Why is that?

 

[Doc Al]

Hint: θ is not the angle of incidence.

 

[HunterDX77M]

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. . . Thank you.