Question

A model of an optical fiber is shown in the figure below. The optical fiber has an index of refraction, $n$, and is surrounding by free space. What angles of incidence, $\theta$, will result in the light being totally internally reflected at the second surface?

• A. $\theta > \sin^{-1}(\sqrt{n^2-1})$
• B. $\theta < \sin^{-1}(\sqrt{n^2-1})$
• C. $\theta < \sin^{-1}(\sqrt{n^2+1})$
• D. $\sin^{-1}(\sqrt{n^2-1}) < \theta < \sin^{-1}(\sqrt{n^2+1})$

Answer 1

Answer: (B)

(B)

Answer 2

To determine the angles of incidence, $\theta$, that result in total internal reflection (TIR) at the second surface of the optical fiber, we need to apply Snell's Law and the condition for TIR.

1. Refraction at the first surface:

Let the angle of incidence from free space (refractive index $n_0=1$) into the optical fiber (refractive index $n$) be $\theta$. Let the angle of refraction inside the fiber be $r$.

According to Snell's Law at the first surface:

$n_0 \sin \theta = n \sin r$

$1 \cdot \sin \theta = n \sin r$

$\sin r = \frac{\sin \theta}{n}$ (Equation 1)

2. Geometry of angles:

From the figure, the first surface is vertical and the second surface is horizontal. The normal to the first surface is horizontal, and the normal to the second surface is vertical. The refracted ray inside the fiber connects these two surfaces.

Let $i_c$ be the angle of incidence at the second surface. From the geometry (considering the right-angled triangle formed by the ray and the two normals), the angle of refraction $r$ and the angle of incidence $i_c$ are complementary:

$r + i_c = 90^\circ$

So, $i_c = 90^\circ - r$ (Equation 2)

3. Condition for Total Internal Reflection (TIR) at the second surface:

For TIR to occur at the interface between the optical fiber (index $n$) and free space (index $1$), the angle of incidence $i_c$ must be greater than or equal to the critical angle $\theta_c$.

The critical angle $\theta_c$ is defined by:

$n \sin \theta_c = 1 \cdot \sin 90^\circ$

$n \sin \theta_c = 1$

$\sin \theta_c = \frac{1}{n}$

So, the condition for TIR is:

$i_c \ge \theta_c$

Taking the sine of both sides (since $i_c$ and $\theta_c$ are acute angles):

$\sin i_c \ge \sin \theta_c$

$\sin i_c \ge \frac{1}{n}$ (Equation 3)

4. Substitute and solve for $\theta$:

Substitute $i_c = 90^\circ - r$ from Equation 2 into Equation 3:

$\sin(90^\circ - r) \ge \frac{1}{n}$

$\cos r \ge \frac{1}{n}$

Now, we need to express $\cos r$ in terms of $\sin \theta$. We know that $\cos^2 r + \sin^2 r = 1$, so $\cos r = \sqrt{1 - \sin^2 r}$ (since $r$ is an angle of refraction, $0^\circ \le r \le 90^\circ$, so $\cos r \ge 0$).

Substitute $\sin r = \frac{\sin \theta}{n}$ from Equation 1:

$\cos r = \sqrt{1 - \left(\frac{\sin \theta}{n}\right)^2} = \sqrt{1 - \frac{\sin^2 \theta}{n^2}}$

Substitute this back into the TIR condition:

$\sqrt{1 - \frac{\sin^2 \theta}{n^2}} \ge \frac{1}{n}$

Square both sides (both sides are positive):

$1 - \frac{\sin^2 \theta}{n^2} \ge \frac{1}{n^2}$

Multiply by $n^2$:

$n^2 - \sin^2 \theta \ge 1$

Rearrange the inequality to solve for $\sin^2 \theta$:

$n^2 - 1 \ge \sin^2 \theta$

$\sin^2 \theta \le n^2 - 1$

Take the square root of both sides. Since $\theta$ is an angle of incidence ($0^\circ \le \theta \le 90^\circ$), $\sin \theta \ge 0$:

$\sin \theta \le \sqrt{n^2 - 1}$

Finally, to find the angle $\theta$:

$\theta \le \sin^{-1}(\sqrt{n^2 - 1})$

This condition means that for total internal reflection to occur, the angle of incidence $\theta$ must be less than or equal to $\sin^{-1}(\sqrt{n^2 - 1})$.

Comparing this with the given options, option (B) is $\theta < \sin^{-1}(\sqrt{n^2 - 1})$. While our derivation includes the equality (meaning TIR also occurs at the critical angle where the ray grazes the surface), option (B) represents the range where the light is clearly reflected back into the fiber. In multiple-choice questions, the strict inequality is often used to imply clear reflection rather than grazing. Given the options, (B) is the correct choice. Options (A), (C), and (D) are incorrect based on the derivation.