Question

a beam of light enters the end of an optic fiber as shown in (figure 1)
figure
part a
show that we can guarantee total internal reflection at the side surface of the material (at point a), if the index of refraction is greater than about 1.41. in other words, regardless of the angle α, the light beam reflects back into the material at point a, assuming air outside. what if the fiber were immersed in water?
essay answers are limited to about 500 words (3800 characters maximum, including spaces).
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Explanation:

To guarantee total internal reflection (TIR) at point A (side surface of the optic fiber material), we use Snell’s law and TIR conditions.

Step 1: Snell’s Law at Entry

When light enters the material from air, Snell’s law gives:
$$n_{\text{air}} \sin\alpha = n \sin\beta$$
where \( n_{\text{air}} \approx 1 \), \( n \) is the material’s refractive index, \( \alpha \) is the angle of incidence in air, and \( \beta \) is the angle of refraction in the material.

Step 2: Geometry at the Side Surface

At the side surface (point A), the angle of incidence for TIR is \( \gamma \). From geometry, \( \beta + \gamma = 90^\circ \), so \( \gamma = 90^\circ - \beta \), and \( \sin\gamma = \cos\beta \).

Step 3: TIR Condition

TIR occurs when \( \sin\gamma \geq \frac{n_{\text{outside}}}{n} \) (where \( n_{\text{outside}} \) is the refractive index of the outside medium, e.g., air \( n_{\text{air}} = 1 \) or water \( n_{\text{water}} \approx 1.33 \)). Substituting \( \sin\gamma = \cos\beta \), the condition becomes:
$$\cos\beta \geq \frac{n_{\text{outside}}}{n}$$

Step 4: Maximize \( \alpha \) (Critical Case)

To guarantee TIR for any \( \alpha \), we consider the maximum \( \alpha \) (when \( \alpha = 90^\circ \), so \( \sin\alpha = 1 \)). From Snell’s law, \( \sin\beta = \frac{1}{n} \) (since \( \sin\alpha = 1 \)). Then, \( \cos\beta = \sqrt{1 - \sin^2\beta} = \sqrt{1 - \frac{1}{n^2}} \).

Step 5: Solve for \( n \) (Air Outside)

Substitute \( \cos\beta \) into the TIR condition:
$$\sqrt{1 - \frac{1}{n^2}} \geq \frac{1}{n}$$
Square both sides (both sides positive):
$$1 - \frac{1}{n^2} \geq \frac{1}{n^2}$$
$$1 \geq \frac{2}{n^2}$$
$$n^2 \geq 2$$
$$n \geq \sqrt{2} \approx 1.414$$

Step 6: Water Outside (Bonus)

If the fiber is in water (\( n_{\text{water}} \approx 1.33 \)), repeat the critical case (\( \alpha = 90^\circ \), so \( \sin\beta = \frac{n_{\text{water}}}{n} \)). Then \( \cos\beta = \sqrt{1 - \frac{n_{\text{water}}^2}{n^2}} \), and the TIR condition is:
$$\sqrt{1 - \frac{n_{\text{water}}^2}{n^2}} \geq \frac{n_{\text{water}}}{n}$$
Squaring:
$$1 - \frac{n_{\text{water}}^2}{n^2} \geq \frac{n_{\text{water}}^2}{n^2}$$
$$1 \geq \frac{2n_{\text{water}}^2}{n^2}$$
$$n^2 \geq 2n_{\text{water}}^2$$
$$n \geq n_{\text{water}}\sqrt{2} \approx 1.33 \times 1.414 \approx 1.87$$

Summary

To guarantee TIR at the side surface for any \( \alpha \), the material’s refractive index \( n \) must be at least \( \sqrt{2} \approx 1.41 \) (when outside is air). This ensures that even when \( \alpha \) is maximum (90°), the angle of incidence at the side surface (\( \gamma \)) exceeds the critical angle for TIR.

Answer:

To guarantee total internal reflection (TIR) at the side surface (point A) for any angle \( \boldsymbol{\alpha} \), we analyze the geometry, Snell’s law, and TIR conditions:

1. Snell’s Law at Entry (Air → Material)

When light enters the optic fiber (refractive index \( n \)) from air (\( n_{\text{air}} = 1 \)), Snell’s law gives:
$$n_{\text{air}} \sin\alpha = n \sin\beta \implies \sin\beta = \frac{\sin\alpha}{n}.$$

2. Geometry at the Side Surface

At point A, the angle of incidence for TIR is \( \boldsymbol{\gamma} \). From the right triangle (see Figure), \( \beta + \gamma = 90^\circ \), so \( \gamma = 90^\circ - \beta \). Thus, \( \sin\gamma = \cos\beta \).

3. TIR Condition

TIR occurs when the angle of incidence \( \gamma \) satisfies:
$$\sin\gamma \geq \frac{n_{\text{outside}}}{n},$$
where \( n_{\text{outside}} \) is the refractive index of the medium outside the fiber (air: \( n_{\text{air}} = 1 \); water: \( n_{\text{water}} \approx 1.33 \)). Substituting \( \sin\gamma = \cos\beta \), the condition becomes:
$$\cos\beta \geq \frac{n_{\text{outside}}}{n}.$$

4. Critical Case: Maximum \( \boldsymbol{\alpha} \)

To guarantee TIR for any \( \alpha \), we consider the largest possible \( \alpha \) (when \( \alpha = 90^\circ \), so \( \sin\alpha = 1 \)). From Snell’s law:
$$\sin\beta = \frac{1}{n}.$$
Using the Pythagorean identity, \( \cos\beta = \sqrt{1 - \sin^2\beta} = \sqrt{1 - \frac{1}{n^2}} \).

5. Solve for \( \boldsymbol{n} \) (Air Outside)

Substitute \( \cos\beta \) into the TIR condition:
$$\sqrt{1 - \frac{1}{n^2}} \geq \frac{1}{n}.$$
Square both sides (valid since both sides are positive):
$$1 - \frac{1}{n^2} \geq \frac{1}{n^2} \implies 1 \geq \frac{2}{n^2} \implies n^2 \geq 2 \implies n \geq \sqrt{2} \approx 1.414.$$

6. Water Outside (Bonus)

If the fiber is immersed in water (\( n_{\text{water}} \approx 1.33 \)), repeat the critical case (\( \alpha = 90^\circ \), so \( \sin\beta = \frac{n_{\text{water}}}{n} \)). Then:
$$\cos\beta = \sqrt{1 - \frac{n_{\text{water}}^2}{n^2}},$$
and the TIR condition becomes:
$$\sqrt{1 - \frac{n_{\text{water}}^2}{n^2}} \geq \frac{n_{\text{water}}}{n}.$$
Squaring and simplifying:
$$1 - \frac{n_{\text{water}}^2}{n^2} \geq \frac{n_{\text{water}}^2}{n^2} \implies 1 \geq \frac{2n_{\text{water}}^2}{n^2} \implies n \geq n_{\text{water}}\sqrt{2} \approx 1.87.$$

Conclusion

• For air outside, a refractive index \( \boldsymbol{n \geq \sqrt{2} \approx 1.41} \) guarantees TIR at the side surface for any \( \alpha \).
• For water outside, \( \boldsymbol{n \geq \approx 1.87} \) is required.

This works because the critical angle for TIR is minimized (or guaranteed) when \( n \) is sufficiently large, ensuring \( \gamma \) always exceeds the critical angle.