# Total internal refraction inside an optical fiber

1. Aug 3, 2012

### HunterDX77M

1. The problem statement, all variables and given/known data

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?

2. Relevant equations
Snell's Law:
$$n_1 sin(\theta_1) = n_2 sin(\theta_2)$$

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

3. 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, θc. 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?

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2. Aug 3, 2012

### Staff: Mentor

Hint: θ is not the angle of incidence.

3. Aug 3, 2012

### HunterDX77M

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. . . Thank you.