Chapter 2, Problem 34P

To determine

Difference in the speed of the light flowing along with the direction of flow of water and opposite to the direction of the flow of the water.

Answer to Problem 34P

Difference in the speed of the light flowing along with the direction of flow of water and opposite to the direction of the flow of the water is $4.35\text{ m / s}$.

Explanation of Solution

Write the velocity of the light flowing in the direction of flow of water. Thus,

$\begin{array}{l}{u}_1=\frac{u\prime +v}{1+u\prime v/c^2}\\ =\frac{c/n+v}{1+v/nc}\\ =\frac{c}{n}\frac{\left(1+\frac{nv}{c}\right)}{\left(1+\frac{v}{nc}\right)}\end{array}$

Where, v is the velocity of the flow of the water, n is the index refraction of the water, c is the speed of the light and $u\prime =\frac{c}{n}$ is the speed of the light in the medium with index refraction n.

Further simplify the above equation by using binomial expansion because $v\ll c$. Thus,

$\begin{array}{l}{u}_1=\frac{c}{n}\left(1+\frac{nv}{c}\right){\left(1+\frac{v}{nc}\right)}^{-1}\\ \text{ =}\frac{c}{n}\left(1+\frac{nv}{c}\right)\left(1-\frac{v}{nc}\right)\\ \text{= }\frac{c}{n}\left(1-\frac{v}{nc}+\frac{nv}{c}-\frac{v^2}{c^2}\right)\end{array}$

Because $v\ll c$, therefore, $\frac{v^2}{c^2}\to 0$, simplify the above equation. Thus,

$u_1\text{= }\frac{c}{n}\left(1-\frac{v}{nc}+\frac{nv}{c}\right)$        (I)

Similarly, write the velocity of the light flowing in opposite direction of flow of water. Thus,

$\begin{array}{l}{u}_2=\frac{u\prime +\left(-v\right)}{1+u\prime \left(-v\right)/c^2}\\ =\frac{c/n-v}{1-v/nc}\\ =\frac{c}{n}\frac{\left(1-\frac{nv}{c}\right)}{\left(1-\frac{v}{nc}\right)}\end{array}$

Further simplify the above equation by using binomial expansion because $v\ll c$. Thus,

$\begin{array}{l}{u}_2=\frac{c}{n}\left(1-\frac{nv}{c}\right){\left(1-\frac{v}{nc}\right)}^{-1}\\ \text{ =}\frac{c}{n}\left(1-\frac{nv}{c}\right)\left(1+\frac{v}{nc}\right)\\ \text{= }\frac{c}{n}\left(1+\frac{v}{nc}-\frac{nv}{c}-\frac{v^2}{c^2}\right)\end{array}$

Because $v\ll c$, therefore, $\frac{v^2}{c^2}\to 0$, simplify the above equation. Thus,

$u_2\text{= }\frac{c}{n}\left(1+\frac{v}{nc}-\frac{nv}{c}\right)$        (II)

Subtract equation (I) from equation (II) to get the difference in the speed of the light flowing along with the direction of flow of water and opposite to the direction of the flow of the water. Thus,

$\begin{array}{l}\Delta u=u_1-u_2\\ \text{= }\frac{c}{n}\left(1-\frac{v}{nc}+\frac{nv}{c}\right)-\frac{c}{n}\left(1+\frac{v}{nc}-\frac{nv}{c}\right)\\ \text{= }\frac{c}{n}\left(1-\frac{v}{nc}+\frac{nv}{c}-1-\frac{v}{nc}+\frac{nv}{c}\right)\\ =\frac{c}{n}\left(\frac{2nv}{c}-\frac{2v}{nc}\right)\end{array}$

Further simplify the above equation. Thus,

$\Delta u=2v\left(1-\frac{1}{n^2}\right)$

Substitute $v=5\text{ m / s}$ E and $n=1.33$ in the above equation. Thus,

$\begin{array}{c}\Delta u=2\times 5\text{ m / s}\left(1-\frac{1}{{1.33}^2}\right)\\ \text{=10 m / s}\times 0.4346\\ \approx 4.35\text{ m / s }\end{array}$

Conclusion:

Therefore, difference in the speed of the light flowing along with the direction of flow of water and opposite to the direction of the flow of the water is $4.35\text{ m / s}$.