Question

(a) A particular silica glass waveguide is reported to have a loss coefficient of$\mathbf{0}\mathbf{.}\mathbf{050}\mathbf{}\mathbf{dB}\mathbf{/}\mathbf{cm}$(power out/power in, defined in Problem 20-22) for 514-nm-wavelength light. The thickness of the waveguide is $\mathbf{0}\mathbf{.}\mathbf{60}\mathbf{\mu m}$and the length is 3.0 cm. The angle of incidence$\mathbf{(}{\mathit{\theta }}_{\mathbf{i}}$in the figure ) $\mathbf{70}\mathbf{°}$is . What fraction of the incident radiant intensity is transmitted through the waveguide?

(b) If the index of refraction of the waveguide material is 1.5, what is the wavelength of light inside the waveguide? What is the frequency?

Short answer

(a) The power out/power quotient is 0.9649.

(b) The light inside the waveguide has a wavelength is 343 nm and the light inside the waveguide has a frequency is$5.83\times {10}^{14}\mathrm{Hz}$

Step-by-step solution

Definition of waveguide and frequency

• A Waveguide is a hollow metallic tube with a consistent cross section for transmitting electromagnetic waves by successive reflections from the tube's inner walls.
• The number of waves that travel through a particular spot in a given amount of time is referred to as frequency.
• It also refers to the number of cycles or vibrations a body in periodic motion goes through in one unit of time.

Determine the power out/power quotient

(a)

It is necessary to calculate the power out/power quotient.

The following formula can be used to determine the quotient power out/power.

$\frac{\mathrm{Powerout}}{\mathrm{Powerin}}={10}^{-1}\left(\mathrm{dB}/\mathrm{m}/10\right)$

$\mathrm{dB}/\mathrm{m}=$decibel per meter

I=an optical fiber of length$\left(0.60\mathrm{\mu m}\right)$

Given,

$\mathrm{l}=\mathrm{a}$length of optical fiber $\left(0.60\mathrm{\mu m}\right)$

A decibel per metre is a measurement of how loud something is 0.0100 dB/m

The path of a light wave through a waveguide can be estimated as shown in the diagram.

The number of length intervals $\mathrm{lin}3.0\mathrm{cm}$is $\frac{\left(3.0\mathrm{cm}\right)}{\left(1.648\mathrm{\mu m}\right)}=1.820\times {10}^4$

As a result, the light's journey length is $\mathrm{h}\times (1.820\times {10}^4)=3.19\mathrm{cm}$

$$\begin{gathered} \frac{\mathrm{Power}\mathrm{out}}{\mathrm{Powerin}}={10}^{-1\left(\mathrm{dB}/\mathrm{m}10\right)} \\ ={10}^{-\left(3.19\mathrm{cm}\right)\left(0.050\mathrm{dB}/\mathrm{m}10\right)} \\ =0.9649 \end{gathered}$$

The power out/power quotient is 0.9649.

Determine the wavelength of light inside the waveguide and the frequency

(b)

The wavelength and frequency of light inside the waveguide must be estimated.

The formula below is used to compute the wavelength and frequency.

Wavelength$=\frac{{\lambda }_0}{n}$

where

${\lambda }_0=$is the wavelength in vacuum

n is the rerfactive index is a measure of how effective a product

Frequency

$v=\frac{c}{\lambda }$

The wavelength of light inside a waveguide can be estimated using the formula:

$$\begin{gathered} \mathrm{Wavelength}=\frac{{\mathrm{\lambda }}_0}{\mathrm{n}} \\ =\frac{514\mathrm{nm}}{1.5} \\ =343\mathrm{nm} \end{gathered}$$

The light inside the waveguide has a wavelength of 343 nm

Frequency

$v=\frac{c}{\lambda }$

Given,

$$\begin{gathered} \mathrm{c}=2.9979\times {10}^8\mathrm{m}/\mathrm{s} \\ \mathrm{\lambda }=514\times {10}^{-9}\mathrm{m} \\ \mathrm{v}=\frac{2.9979\times {10}^3\mathrm{m}/\mathrm{s}}{514\times {10}^{14}\mathrm{Hz}} \\ \mathrm{v}=5.83\times {10}^{14}\mathrm{Hz} \end{gathered}$$

The light frequency inside the waveguide is$5.83\times {10}^{14}\mathrm{Hz}$