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Expert-verified Found in: Page 1250 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # The active medium in a particular laser generates laser light at a wavelength of 694nm is 6.00 cm long and 1.00 cm in diameter. (a) Treat the medium as an optical resonance cavity analogous to a closed organ pipe. How many standing-wave nodes are there along the laser axis? (b) By what amount ${\mathbf{∆}}{\mathbf{f}}$ would the beam frequency have to shift to increase this number by one? (c) Show that ${\mathbf{∆}}{\mathbf{f}}$ is just the inverse of the travel time of laser light for one round trip back and forth along the laser axis. (d) What is the corresponding fractional frequency shift ${\mathbf{∆}}{\mathbf{f}}{\mathbf{/}}{\mathbf{f}}$ ? The appropriate index of refraction of the lasing medium (a ruby crystal) is 1.75.

(a) There are $3.03×{10}^{5}$ standing wave nodes along the laser axis.

(b) The beam frequency needs to shift by $1.43×{10}^{9}\mathrm{Hz}$ to increase the nodes value by 1.

(c) The shift in beam frequency $∆f$ is just the inverse of the travel time of laser light for one round trip back and forth along the laser axis.

(d) The corresponding fractional frequency shift $∆f/f$ is $3.31×{10}^{-6}$ .

See the step by step solution

## Step 1: The given data:

The wavelength of the laser, $\lambda =694\mathrm{c}\mathrm{nm}$

Length of the laser, L= 6 cm

Diameter of the laser, d =1 cm

The appropriate index of refraction of the lasing medium is n =1.75 .

## Step 2: Understanding the concept of standing waves and frequency:

A standing wave is a combination of two waves moving in opposite directions, each with the same amplitude and frequency.

If both the mirrors are perfectly reflecting, there is a node at each end of the crystal. Thus, with one end partially silvered, there is a node very close to that end. Now, assume their nodes at both ends, then there are an integral number of half-wavelength in the length of the crystal. Thus, considering the wavelength due to refraction gives the number of nodes by using the standing wave equation.

Now, for the frequency shift, consider the frequency equation considering the condition that the nodes number is increased by 1. Now, similarly, the speed is calculated using the same approach. And for the fractional change, calculate the normal frequency to divide the frequency shift.

Formula:

The standing wave equation is,

$N\lambda =2L$ ….. (1)

The wavelength of a light source is,

$\lambda =\frac{C}{f}$ ….. (2)

## Step 3: (a) Calculation of the number of nodes:

Now, for the case of refraction, the wavelength becomes, ${\lambda }_{c}=\lambda }{n}$

Now, substituting this value in the equation (1) and using the given data, the number of standing wave nodes can be calculated as follows:

$N\left(\lambda /n\right)=2L$

$\mathrm{N}=\frac{2\mathrm{Ln}}{\mathrm{\lambda }}\phantom{\rule{0ex}{0ex}}\mathrm{N}=\frac{2\left(0.06\mathrm{m}\right)\left(1.75\right)}{694×{10}^{-9}\mathrm{m}}\phantom{\rule{0ex}{0ex}}=3.03×{10}^{5}$ ….. (3)

Hence, the value of the number of the nodes is $3.03×{10}^{5}$.

## Step 4: (b) Calculation of the frequency shift if nodes value is increased by 1:

To obtain the frequency shift due to increase in node value, we substitute the equation (2) in equation (3) and then differentiate the equation as follows:

$∆\mathrm{N}=\frac{2\mathrm{Ln}∆\mathrm{f}}{\mathrm{c}}\phantom{\rule{0ex}{0ex}}∆\mathrm{f}=\frac{\mathrm{c}∆\mathrm{N}}{2\mathrm{Ln}}\phantom{\rule{0ex}{0ex}}=\frac{\left(3×{10}^{8}\mathrm{m}/\mathrm{s}\right)}{2\left(0.06\mathrm{m}\right)\left(1.75\right)}\phantom{\rule{0ex}{0ex}}=1.43×{0}^{9}\mathrm{Hz}$

Hence, the value of frequency shift is $1.43×{10}^{9}\mathrm{Hz}$.

## Step 5: (c) Calculation of the shift in beam frequency:

Now, the speed of light in the crystal is and the round-trip distance is 2L, so the round-trip travel time is 2nL/c . This is the same as the reciprocal of the change in frequency.

Hence, the shift in beam frequency $∆\mathrm{f}$ is just the inverse of the travel time of laser light for one round trip back and forth along the laser axis.

## Step 6: (d) Calculation of the fractional change in frequency

The frequency of the beam can be calculated using the given data in equation (2) as follows:

$\mathrm{f}=\frac{3×{10}^{8}\mathrm{m}/\mathrm{s}}{694×{10}^{-9}\mathrm{m}}\phantom{\rule{0ex}{0ex}}=4.32×{10}^{14}\mathrm{Hz}$

Thus, the fractional change in frequency is given as follows:

$\frac{∆\mathrm{f}}{\mathrm{f}}=\frac{1.43×{10}^{9}\mathrm{Hz}}{4.32×{10}^{14}\mathrm{Hz}}\phantom{\rule{0ex}{0ex}}=3.31×{10}^{-6}$

Hence, the value of the fractional change is $3.31×{10}^{-6}$. ### Want to see more solutions like these? 