Chapter 35, Problem 49P

(a)

To determine

The smallest outside radius permitted for a bend in the fiber.

Answer to Problem 49P

The smallest outside radius permitted for a bend in the fiber is $\frac{nd}{n-1}$.

Explanation of Solution

The ray diagram for the optical fiber is shown in figure below.

Figure (1)

It is given that,

    $n_1=n$

Here, $n_1$ is the refractive index of medium $1$ and $n$ is the refractive index of optical fiber.

Write the expression to calculate the critical angle.

    $\sin {\theta }_{\text{c}}=\frac{n_2}{n_1}$                                                                                                                  (I)

Here, ${\theta }_{\text{c}}$ is the critical angle and $n_2$ is the refractive index of air.

Re arrange equation (I) to get the expression for ${\theta }_{\text{c}}$.

    ${\theta }_{\text{c}}={\sin }^{-1}\left(\frac{n_2}{n_1}\right)$

Consider figure (1).

Since the light ray on the left side of point $A$ is incident on the fiber tangentially, so the incident angle is $90°$. Hence, total internal reflection will take place.

From figure (1).

    $BC=AC-AB$                                                                                                         (II)

Substitute $R$ for $AC$ and $d$ for $AB$

    $BC=R-d$

Here, $R$ is the radius of curvature of the fiber and $d$ is the diameter of the fiber.

Also,

    $DC=R$

Since, $DC$ is perpendicular to the tangent $GF$ at $D$. Hence, the angle of incidence of the ray $BD$ at $D$ is,

    $\angle BDC=\theta$

Consider $\Delta BCD$.

    $\frac{BC}{CD}=\sin \theta$                                                                                                              (III)

Substitute $R-d$ for $BC$ and $R$ for $CD$ in equation (III).

    $\begin{array}{c}\frac{R-d}{R}=\sin \theta \\ \theta ={\sin }^{-1}\left(\frac{R-d}{R}\right)\end{array}$                                                                                             (IV)

The condition for total internal reflection to occur is,

    $\theta \ge {\theta }_{\text{c}}$                                                                                                                            (V)

Substitute ${\sin }^{-1}\left(\frac{n_2}{n_1}\right)$ for ${\theta }_{\text{c}}$ and ${\sin }^{-1}\left(\frac{R-d}{R}\right)$ for $\theta$ in equation (V).

    $\begin{array}{c}{\sin }^{-1}\left(\frac{R-d}{R}\right)\ge {\sin }^{-1}\left(\frac{n_2}{n_1}\right)\\ \left(\frac{R-d}{R}\right)\ge \left(\frac{n_2}{n_1}\right)\\ n_{1}R-n_{1}d\ge n_{2}R\\ R\left(n_1-n_2\right)\ge n_{1}d\end{array}$                                                                                           (VI)

Conclusion:

Substitute $1$ for $n_2$ and $n$ for $n_1$ in equation (VI) to solve for $R$.

    $\begin{array}{c}R\left(n-1\right)\ge nd\\ R\ge \frac{nd}{\left(n-1\right)}\end{array}$                                                                                                       (VII)

So the smallest outside radius for no light to escape is,

    $R_{\min }=\frac{nd}{n-1}$                                                                                                                  (VIII)

Therefore, the smallest outside radius permitted for a bend in the fiber is $\frac{nd}{n-1}$.

(b).

To determine

The effect on the smallest outer radius when $d$ approaches $0$ and also explain if this behavior reasonable.

Answer to Problem 49P

The effect on the smallest outer radius when $d$ approaches $0$ is $R_{\min }\to 0$. This behavior is reasonable because for very small diameter, the light strikes the interface at very large angles of incidence.

Explanation of Solution

Rearrange the expression for $R_{\min }$ as,

    $R_{\min }=\frac{d}{1-\frac{1}{n}}$                                                                                                          (IX)

From equation (IX), it is clearly visible that $d$ has a direct relationship with $R_{\min }$.

So, as $d$ approaches $0$, then $R_{\min }$ will also approach to $0$.

    $R_{\min }\to 0$

Yes, this behavior is reasonable because for very small diameter, the light strikes the interface at very large angles of incidence.

(c).

To determine

The effect on the smallest outer radius when $n$ increases and also explain if this behavior reasonable.

Answer to Problem 49P

The value of smallest outer radius $R_{\min }$ decreases as $n$ increases. This behavior is reasonable because as $n$ increases, the critical angle becomes smaller.

Explanation of Solution

From equation (IX), when $n$ increases the denominator terms keeps on increasing. Thereby the value of $R_{\min }$ decreases.

From equation (I), when $n$ increases the critical angle decreases.

Thus, the above result is reasonable because with the increase in the value of $n$, the critical angle decreases.

(d).

To determine

The effect on the smallest outer radius when $n$ approaches unity and also explain if this behavior reasonable.

Answer to Problem 49P

The value of smallest outer radius $R_{\min }$ approaches to infinity as $n$ approaches $1$. This result is reasonable because the critical angle becomes close to $90°$ and any bend will allow the light to escape.

Explanation of Solution

From equation (IX), when $n$ approaches to unity, then the denominator is $0$, thereby the whole equation become infinity that is the value of $R_{\min }$ tends to infinity.

    $R_{\min }\to \infty$

From equation (I), when $n$ approaches to unity, the critical angle approaches to $90°$.

This result is reasonable because the critical angle becomes close to $90°$ and any bend will allow the light to escape.

(e).

To determine

The value of $R_{\min }$

Answer to Problem 49P

The value of $R_{\min }$ is $350\mu \text{m}$.

Explanation of Solution

Conclusion:

Substitute $100\mu \text{m}$ for $d$ and $1.40$ for $n$ in equation (VIII) to solve for $R_{\min }$.

    $\begin{array}{c}{R}_{\min }=\frac{1.40\left(100\mu \text{m}\right)}{1.40-1}\\ =350\mu \text{m}\end{array}$

Therefore, the value of $R_{\min }$ is $350\mu \text{m}$.