Question

The active medium in a particular laser that generates laser light at a wavelength of 694 nm 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 $\Delta f$ would the beam frequency have to shift to increase this number by one? (c) Show that $\Delta f$ 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 $\dfrac{{\Delta f}}{f}$​? The appropriate index of refraction of the lasing medium (a ruby crystal) is 1.75.

Answer 1

Hint: In this question we have to apply the concept of waves. We have to use the concept of closed organ pipe for the first part. Then we have to calculate how much frequency change is required to increases the number of nodes by one. In the third part we have to prove the fact that $\Delta f$ is just the inverse of the travel time of laser light for one round trip back and forth along the laser axis. And after that we have to calculate the fractional frequency shift corresponding to the change in number of nodes.

Complete step by step answer:A) We are given the wavelength of the laser beam to be 694 nanometers.
Now we have to consider the medium as an optical resonance cavity. Which is analogous to a closed organ pipe.
If we consider the formula of number of nodes $N$ in a closed organ pipe it is:
$L = \dfrac{{N{\lambda _C}}}{2}$,
Where $L$ is the length of the active medium.
${\lambda _C}$ is the wavelength of the ray inside the active medium.
The formula for ${\lambda _C}$ is given by:
${\lambda _C} = \dfrac{\lambda }{\mu }$, where $\lambda $is the Wavelength of ray outside the active medium and $\mu $ is the refractive index of the active medium.
If we put this value of ${\lambda _C}$ in the formula of number of nodes $N$ and make $N$ the subject, we get:
$\begin{array}{l}
L = \dfrac{{N{\lambda _c}}}{2}\\
\Rightarrow L = \dfrac{{N\lambda }}{{2\mu }}\\
\Rightarrow N = \dfrac{{2\mu L}}{\lambda }
\end{array}$


Now we put the value given in the question with proper units conversion,
$\begin{array}{l}
N = \dfrac{{2 \times 6 \times {{10}^{ - 2}} \times 1.75}}{{694 \times {{10}^{ - 9}}}}\\
\Rightarrow N = \dfrac{{2 \times 6 \times {{10}^{ - 2}} \times 1.75}}{{694 \times {{10}^{ - 9}}}}\\
\Rightarrow N = 3.03 \times {10^5}
\end{array}$
Therefore, the number of nodes is equal to $3.03 \times {10^5}$.

B) Now the condition to be tested is that if the change in the number of nodes is 1 then how much the shift in frequency will be.
We know that $c = \lambda f$, then $\lambda = \dfrac{c}{f}$ . And if we put this value of $\lambda $ in the formula of $N$, we get:
$\begin{array}{l}
N = \dfrac{{2L\mu }}{\lambda }\\
\Rightarrow N = \dfrac{{2Lf\mu }}{c}
\end{array}$
Now if we make $f$ the subject in the above equation and then find the change in $f$, we get:
$\begin{array}{l}
N = \dfrac{{2Lf\mu }}{c}\\
\Rightarrow f = \dfrac{{cN}}{{2L\mu }}\\
\Delta f = \dfrac{{c \times \Delta N}}{{2L\mu }}
\end{array}$
Now we know that $\Delta N = 1$ and putting this value in the formula and calculating we get:
$\begin{array}{l}
\Delta f = \dfrac{{c \times \Delta N}}{{2L\mu }}\\
\Rightarrow \Delta f = \dfrac{{3 \times {{10}^8} \times 1}}{{2 \times 6 \times {{10}^{ - 2}} \times 1.75}}\\
\Rightarrow \Delta f = \dfrac{{3 \times {{10}^8} \times 1}}{{2 \times 6 \times {{10}^{ - 2}} \times 1.75}}\\
\Rightarrow \Delta f = 1.43 \times {10^9}\,Hz
\end{array}$

Hence the shift in frequency is equal to $1.43 \times {10^9}Hz$.
C) In this part we have to prove the relation, $\Delta f = \dfrac{1}{t}$.
We know that the velocity od ray inside the medium is given by,
$v = \dfrac{c}{\mu }$, where c is speed of light in free space.
We also know that,
 $\begin{array}{l}
v = \dfrac{d}{t}\\
\Rightarrow v = \dfrac{{2L}}{t}
\end{array}$
$t = \dfrac{{2L}}{v}$.
Here distance travelled is taken as 2L because the motion to be considered is of back and forth.
If we put the value of v in the above formula, we get:
\[\begin{array}{l}
t = \dfrac{{2L}}{v}\\
\Rightarrow t = \dfrac{{2L\mu }}{c}
\end{array}\]
If we considered the value of change in frequency we know,
$\Delta f = \dfrac{c}{{2\mu L}}$, when $\Delta N = 1$.
We can easily see that the above equation is just the reciprocal of the equation of $t$.
Hence, proved: $\Delta f = \dfrac{1}{t}$.
D) In this part we have to calculate the fractional shift in the frequency.
The mathematical representation of fractional shift in the frequency is:$\dfrac{{\Delta f}}{f}$.
Now we know that $f = \dfrac{c}{\lambda }$we will put this value of f in the formula of fractional shift in frequency:
$\dfrac{{\Delta f}}{f} = \dfrac{{\Delta f\lambda }}{c}$
We know the values of $\Delta f$, $\lambda $and c and we have to put them in the above equation,
$\begin{array}{l}
\dfrac{{\Delta f}}{f} = \dfrac{{1.43 \times {{10}^9} \times 694 \times {{10}^{ - 9}}}}{{3 \times {{10}^8}}}\\
\Rightarrow \dfrac{{\Delta f}}{f} = \dfrac{{1.43 \times {{10}^9} \times 694 \times {{10}^{ - 9}}}}{{3 \times {{10}^8}}}\\
\Rightarrow \dfrac{{\Delta f}}{f} = 3.308 \times {10^{ - 6}}
\end{array}$
Hence, the fractional shift infrequency is equal to $3.308 \times {10^{ - 6}}$.

Note: In the question students should calculate the required quantities one by one and use them to calculate the quantities in the next parts. If one tries to calculate every quantity from the very beginning then it will consume a lot more time than it should take.