# Rearranging the equation for the cutoff condition in optical fibers

• EmilyRuck
In summary, the conversation discusses the characteristic equation for modes in optical fibers. The equation is derived using normalized quantities and involves Bessel functions. The cutoff condition for modes is obtained by taking the limit of the equation as one of the normalized quantities approaches zero. There is a discrepancy between the equation derived in the conversation and the one in the linked document, but it is resolved by correcting a formula in the document.
EmilyRuck
Hello!

In Optical fibers, let ##k_1## and ##k_2## be respectively the propagation constants in core and cladding, ##\beta## the propagation costant of a mode along the direction ##z##, ##a## the radius of the fiber. Using the normalized quantities ##u=a \sqrt{k_1^2 − \beta^2}## and ##w=a \sqrt{\beta^2 − k_2^2}##, the characteristic equation for modes is:

$$\left[ \displaystyle \frac{J'_{\nu} (u)}{u J_{\nu}(u)} + \frac{K'_{\nu} (w)}{w K_{\nu}(w)} \right] \left[ \displaystyle \frac{k_1^2 J'_{\nu} (u)}{u J_{\nu}(u)} + \frac{k_2^2 K'_{\nu} (w)}{w K_{\nu}(w)} \right] = \nu^2 \beta^2 \left( \displaystyle \frac{1}{u^2} + \frac{1}{w^2} \right)^2$$

where ##n_1## and ##n_2## are the refractive indices of core and cladding, ##\nu## is the order of Bessel functions.

As specified in this document, page 15, the cutoff condition for modes is obtained taking the limit of the above expression for ##w \to 0##. But, before this, I can't obtain the same expression as in the document. Consider:

$$\frac{J'_{\nu} (u)}{u J_{\nu}(u)} = \frac{J_{\nu - 1} (u)}{u J_{\nu}(u)} - \frac{\nu}{u^2} = \xi_1(u) - \frac{\nu}{u^2}\\ \frac{K'_{\nu} (w)}{w K_{\nu}(u)} = \frac{K_{\nu - 1} (w)}{w K_{\nu}(w)} + \frac{\nu}{w^2} = \xi_2(w) + \frac{\nu}{w^2}$$

Substituting in the characteristic equation, dividing both sides by ##k_1^2## and rearranging the terms, I obtain:

$$\xi_1^2(u) + \xi_1(u) \left[ \frac{k_1^2 + k_2^2}{k_1^2} \cdot \xi_2(w) - \nu \left( \frac{2}{u^2} - \frac{k_1^2 + k_2^2}{k_1^2} \cdot \frac{1}{w^2} \right) \right] + \left[ \frac{k_2^2}{k_1^2} \cdot \xi_2^2(w) - \xi_2(w) \nu \left( \frac{k_1^2 + k_2^2}{k_1^2} \cdot \frac{1}{u^2} - 2 \frac{k_2^2}{k_1^2} \cdot \frac{1}{w^2} \right) \right] + \left( \frac{\nu}{u^2} \right)^2 - \frac{\nu^2}{u^2 w^2} - \frac{k_2^2}{k_1^2} \cdot \frac{\nu^2}{u^2 w^2} + \frac{k_2^2}{k_1^2} \left( \frac{\nu}{w^2} \right)^2 - \frac{1}{k_1^2} \left( \frac{\nu^2 \beta^2}{u^4} + 2 \frac{\nu^2 \beta^2}{u^2 w^2} + \frac{\nu^2 \beta^2}{w^4} \right) = 0$$

I double checked this result, but the document shows instead:

$$\xi_1^2(u) - \xi_1(u) \left[ \frac{k_1^2 + k_2^2}{k_1^2} \cdot \xi_2(w) + \nu \left( \frac{2}{u^2} + \frac{k_1^2 + k_2^2}{k_1^2} \cdot \frac{1}{w^2} \right) \right] + \left[ \frac{k_2^2}{k_1^2} \cdot \xi_2^2(w) + \xi_2(w) \nu \left( \frac{k_1^2 + k_2^2}{k_1^2} \cdot \frac{1}{u^2} + 2 \frac{k_2^2}{k_1^2} \cdot \frac{1}{w^2} \right) \right] = 0$$

First, terms with ##\frac{k_1^2 + k_2^2}{k_1^2}## have swapped signs; moreover, I can't figure out how it could be

$$\left( \frac{\nu}{u^2} \right)^2 - \frac{\nu^2}{u^2 w^2} - \frac{k_2^2}{k_1^2} \cdot \frac{\nu^2}{u^2 w^2} + \frac{k_2^2}{k_1^2} \left( \frac{\nu}{w^2} \right)^2 - \frac{1}{k_1^2} \left( \frac{\nu^2 \beta^2}{u^4} + 2 \frac{\nu^2 \beta^2}{u^2 w^2} + \frac{\nu^2 \beta^2}{w^4} \right) = 0$$

Is the result in the document correct? Or am I doing something wrong? As an alternative, I'm also looking for a textbook which deals with the procedure to obtain the cutoff condition.EmilyP. S.

Cylindrical Dielectric Waveguide Modes, E. Snitzer, Journal of the Optical Society of America Vol. 51, Issue 5, pp. 491-498 (1961)

Temporarily putting aside the ##\frac{k_1^2 + k_2^2}{k_1^2}## terms signs, consider the part which should be ##0##. The first 4 terms come from the espansion of the LHS (which involves ##\xi_1##, ##\xi_2##) of the original characteristic equation. The last 3 terms directly come from the RHS of the characteristic equation, divided by ##k_1^2##. Considering the homologous terms:

$$\frac{\nu^2}{u^4} \left( 1 - \frac{\beta^2}{k_1^2} \right) - \frac{\nu^2}{u^2 w^2} \left( \frac{k_1^2 + k_2^2}{k_1^2} + 2 \frac{\beta^2}{k_1^2} \right) + \frac{\nu^2}{w^4} \left( \frac{k_2^2}{k_1^2} - \frac{\beta^2}{k_1^2} \right) = \\ = \frac{\nu^2}{u^4} \left( \frac{k_1^2 - \beta^2}{k_1^2} \right) - \frac{\nu^2}{u^2 w^2} \left( \frac{k_1^2 + k_2^2 + 2\beta^2}{k_1^2} \right) + \frac{\nu^2}{w^4} \left( \frac{k_2^2 - \beta^2}{k_1^2} \right)$$

Now, ##u = a k_{t_1}## and ##w = a |k_{t_2}|##. ##k_{t_1}## is, in the core, the wavevector component orthogonal to the direction of propagation: the transverse wavevector; ##k_{t_2}## is its homologous in the cladding, but for a confined mode it is pure imaginary, therefore ##k_{t_2}^2 = - |k_{t_2}|^2##. Then,

$$k_1^2 - \beta^2 = k_{t_1}^2 = \frac{u^2}{a^2}\\ k_2^2 - \beta^2 = - |k_{t_2}|^2 = - \frac{w^2}{a^2}\\ k_{t_1}^2 - |k_{t_2}|^2 = k_1^2 + k_2^2 - 2 \beta^2 \Rightarrow k_1^2 + k_2^2 - 2 \beta^2 = \frac{u^2 - w^2}{a^2}$$

Substituting in the above equation:

$$\frac{\nu^2}{a^2 u^2} - \frac{\nu^2}{a^2 w^2} + \frac{\nu^2}{a^2 u^2} - \frac{\nu^2}{u^2 w^2} 4 \beta^2 - \frac{\nu^2}{a^2 w^2} = \\ = 2 \frac{\nu^2}{a^2 u^2} - 2\frac{\nu^2}{a^2 w^2} - 4 \frac{\nu^2}{u^2 w^2} \beta^2 = 2 \frac{\nu^2}{a^2} \cdot \frac{w^2 - u^2}{u^2 w^2} - 4 \frac{\nu^2}{u^2 w^2} \beta^2 = \\ = - 2 \nu^2 \cdot \frac{k_1^2 + k_2^2 - 2 \beta^2}{u^2 w^2} - 4 \frac{\nu^2}{u^2 w^2} \beta^2 = - 2 \nu^2 \cdot \frac{k_1^2 + k_2^2}{u^2 w^2}$$

I hope I did not make errors. But, still, it doesn't seem that this expression can be ##0##.

The first expression, which is correct, is written using formula (A4) of the linked document:

$$\frac{J'_{\nu} (u)}{u J_{\nu}(u)} = \frac{J_{\nu - 1} (u)}{u J_{\nu}(u)} - \frac{\nu}{u^2} = \xi_1(u) - \frac{\nu}{u^2}$$

Formula (A6), used for the second expression, is wrong. It should be:

$$\frac{K'_{\nu} (w)}{w K_{\nu}(u)} = - \frac{K_{\nu - 1} (w)}{w K_{\nu}(w)} - \frac{\nu}{w^2}$$

And therefore the second expression should be:

$$\frac{K'_{\nu} (w)}{w K_{\nu}(u)} = - \frac{K_{\nu - 1} (w)}{w K_{\nu}(w)} - \frac{\nu}{w^2} = - \xi_2(w) - \frac{\nu}{w^2}$$

Substituting in the characteristic equation, the mentioned issues are no more present. In fact, signs are correct and the extra-term becomes:

$$\left( \frac{\nu}{u^2} \right)^2 + \frac{\nu^2}{u^2 w^2} + \frac{k_2^2}{k_1^2} \cdot \frac{\nu^2}{u^2 w^2} + \frac{k_2^2}{k_1^2} \left( \frac{\nu}{w^2} \right)^2 - \frac{1}{k_1^2} \left( \frac{\nu^2 \beta^2}{u^4} + 2 \frac{\nu^2 \beta^2}{u^2 w^2} + \frac{\nu^2 \beta^2}{w^4} \right)$$

Consider the homologous terms, it becomes:

$$\frac{\nu^2}{u^4} \left( 1 - \frac{\beta^2}{k_1^2} \right) - \frac{\nu^2}{u^2 w^2} \left( - \frac{\left( k_1^2 + k_2^2 \right)}{k_1^2} + 2 \frac{\beta^2}{k_1^2} \right) + \frac{\nu^2}{w^4} \left( \frac{k_2^2}{k_1^2} - \frac{\beta^2}{k_1^2} \right)$$

Substituting

$$- \left( k_1^2 + k_2^2 \right) + 2 \beta^2 = \frac{w^2 - u^2}{a^2}\\ k_1^2 - \beta^2 = \frac{u^2}{a^2}\\ k_2^2 - \beta^2 = - \frac{w^2}{a^2}$$

The whole term vanishes.

The final result is, as expected,

$$\xi_1^2(u) - \xi_1(u) \left[ \frac{k_1^2 + k_2^2}{k_1^2} \cdot \xi_2(w) + \nu \left( \frac{2}{u^2} + \frac{k_1^2 + k_2^2}{k_1^2} \cdot \frac{1}{w^2} \right) \right] + \left[ \frac{k_2^2}{k_1^2} \cdot \xi_2^2(w) + \xi_2(w) \nu \left( \frac{k_1^2 + k_2^2}{k_1^2} \cdot \frac{1}{u^2} + 2 \frac{k_2^2}{k_1^2} \cdot \frac{1}{w^2} \right) \right] = 0$$

## 1. How do you rearrange the equation for the cutoff condition in optical fibers?

The equation for the cutoff condition in optical fibers is typically rearranged to solve for the critical angle of incidence, which is the angle at which light will no longer be transmitted through the fiber. This can be done by setting the refractive index of the core equal to the refractive index of the cladding and solving for the angle.

## 2. What is the significance of the cutoff condition in optical fibers?

The cutoff condition is an important concept in optical fibers because it determines the minimum angle at which light can be transmitted through the fiber. This angle is crucial for maintaining a strong signal and preventing signal loss in the fiber.

## 3. How does the refractive index affect the cutoff condition in optical fibers?

The refractive index of the core and cladding materials plays a crucial role in determining the cutoff condition in optical fibers. A higher refractive index in the core will result in a lower critical angle of incidence, allowing for more efficient transmission of light through the fiber.

## 4. Can the cutoff condition be manipulated to improve fiber optic performance?

Yes, the cutoff condition can be manipulated by adjusting the refractive index of the core and cladding materials. This can be done during the manufacturing process to optimize the performance of the fiber for specific applications.

## 5. Are there any limitations to the cutoff condition in optical fibers?

While the cutoff condition is an important factor in fiber optic performance, it is not the only factor that affects signal transmission. Other factors such as bending, scattering, and losses at connectors can also impact the performance of optical fibers.

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