Chapter 2, Problem 6P

To determine

Velocity of the ether used in the null experiment.

Answer to Problem 6P

The velocity of the ether used in the null experiment is $3.47\times {10}^3\text{ m / s}$.

Explanation of Solution

Number of fringes shifted (n) will be defined as follows:

$n=\frac{\Delta d}{\lambda }$ and $\Delta d=c\left(\Delta t\prime -\Delta t\right)$

The time difference between the complete round trip journeys of the optical lengths before $90°$ rotation of the Michelson interferometer. Thus,

$\begin{array}{l}\Delta t=t_2-t_1\\ =\frac{2}{c}\left(\frac{{\ell }_2}{\sqrt{1-v^2/c^2}}-\frac{{\ell }_1}{1-v^2/c^2}\right)\end{array}$        (I)

Where, ${\ell }_1$ and ${\ell }_2$ are the optical length, $t_1$ is the time taken by the sodium light to complete the round trip from source to target and from target to source along the optical length ${\ell }_1$, $t_2$ is the time taken by the sodium light to complete the round trip from source to target and from target to source along the optical length ${\ell }_2$, c is the speed of the light, and v is the speed of the ether.

The time difference between the complete round trip journeys of the optical lengths after $90°$ rotation of the Michelson interferometer. Thus,

$\begin{array}{l}\Delta t\prime =t_2^{\prime }-t_1^{\prime }\\ =\frac{2}{c}\left(\frac{{\ell }_2}{1-v^2/c^2}-\frac{{\ell }_1}{\sqrt{1-v^2/c^2}}\right)\end{array}$        (II)

Where, c is the speed of the light, $\Delta t\prime$ and $\Delta t$ is the time difference after and before rotating the Michelson interferometer $90°$ respectively, and $\lambda$ is the wavelength of the sodium light.

Where, $t_1^{\prime }$ is the time taken by the sodium light to complete the round trip from source to target and from target to source along the optical length ${\ell }_1$, $t_2^{\prime }$ is the time taken by the sodium light to complete the round trip from source to target and from target to source along the optical length ${\ell }_2$, c is the speed of the light, and v is the speed of the ether.

Substitute the values of the $\Delta t\prime$ and $\Delta t$ in formula $\Delta d=c\left(\Delta t\prime -\Delta t\right)$. Thus,

$\begin{array}{l}\Delta d=c\left(\Delta t\prime -\Delta t\right)\\ =c\left\{\frac{2}{c}\left(\frac{{\ell }_2}{1-v^2/c^2}-\frac{{\ell }_1}{\sqrt{1-v^2/c^2}}\right)\right\}-\left\{\frac{2}{c}\left(\frac{{\ell }_2}{\sqrt{1-v^2/c^2}}-\frac{{\ell }_1}{1-v^2/c^2}\right)\right\}\\ =c\times \frac{2}{c}\left\{\left(\frac{{\ell }_2}{1-v^2/c^2}-\frac{{\ell }_1}{\sqrt{1-v^2/c^2}}\right)-\left(\frac{{\ell }_2}{\sqrt{1-v^2/c^2}}-\frac{{\ell }_1}{1-v^2/c^2}\right)\right\}\\ =2\left\{\left(\frac{{\ell }_2}{1-v^2/c^2}\right)-\left(\frac{{\ell }_1}{\sqrt{1-v^2/c^2}}\right)-\left(\frac{{\ell }_2}{\sqrt{1-v^2/c^2}}\right)+\left(\frac{{\ell }_1}{1-v^2/c^2}\right)\right\}\end{array}$

Further simplify the above equation:

$\Delta d=2\left\{\left(\frac{{\ell }_1+{\ell }_2}{1-v^2/c^2}\right)-\left(\frac{{\ell }_1+{\ell }_2}{\sqrt{1-v^2/c^2}}\right)\right\}$

As $c\gg v$, use binomial expansion to expand the terms involving $\frac{v^2}{c^2}$ and keep only lowest term. Thus,

$\begin{array}{l}\Delta d=2\left({\ell }_1+{\ell }_2\right)\left[\left(1+\frac{v^2}{c^2}+\cdots \right)-\left(1+\frac{v^2}{2c^2}+\cdots \right)\right]\\ \approx \frac{v^2\left({\ell }_1+{\ell }_2\right)}{c^2}\end{array}$        (III)

where l₁ and l₂ are the length of the arm (optical length), which is given as 11 m.

Calculation:

Substitute $\Delta d=\frac{v^2\left({\ell }_1+{\ell }_2\right)}{c^2}$ in $n=\frac{\Delta d}{\lambda }$ and ${\ell }_1$ and ${\ell }_2$ are the length of the arm (optical length), as 11 m. Thus,

$\begin{array}{c}n=\frac{\Delta d}{\lambda }\\ =\frac{v^2\left({\ell }_1+{\ell }_2\right)}{c^2\lambda }\\ v^2=\frac{n{c}^2\lambda }{\left({\ell }_1+{\ell }_2\right)}\\ v=c\sqrt{\frac{n\lambda }{\left({\ell }_1+{\ell }_2\right)}}\end{array}$

Substitute the ${\ell }_1$ and ${\ell }_2$ the length of the arm (optical length), as 11 m, $\lambda =589\text{ nm}=589\times {10}^{-9}\text{ m}$, $n=0.005$, and $c=3.00\times {10}^8\text{ m }/\text{ s}$. Thus,

$\begin{array}{c}v=\left(3\times {10}^8\text{ m }/\text{ s}\right)\sqrt{\frac{(0.005)\left(589\times {10}^{-9}\text{ m}\right)}{\left(11\text{ m}+11\text{ m}\right)}}\\ =\left(3\times {10}^8\text{ m }/\text{ s}\right)\sqrt{133.86\times {10}^{-12}}\\ =3.47\times {10}^3\text{ m }/\text{ s}\end{array}$

Conclusion:

Therefore, the velocity of the ether used in the null experiment is $3.47\times {10}^3\text{ m / s}$