Three observers A , B and C measure the speed of light coming from...

In a vacuum, the speed of light $c$ is invariant for all inertial observers. However, in a refractive medium like water, the speed of light is $v = \dfrac{c}{n} < c$. Because the speed of light in a medium is less than $c$, it is no longer invariant and depends on the observer's frame of reference.

Let the speed of light in water for the stationary observer $B$ be $v_B = v$.

For observer $A$ moving towards the source with speed $u$, the relative speed of light is given by the relativistic velocity addition formula:

$v_A = \dfrac{v + u}{1 + \dfrac{uv}{c^2}}$

Since $v < c$ and $u > 0$, we have $v_A > v$.

For observer $C$ moving away from the source with speed $u$, the relative speed of light is:

$v_C = \dfrac{v - u}{1 - \dfrac{uv}{c^2}}$

This gives $v_C < v$.

Thus, the measured speeds follow the order $v_A > v_B > v_C$.

To check the validity of the relation in option (d), we calculate $v_A + v_C$:

$v_A + v_C = \dfrac{v + u}{1 + \dfrac{uv}{c^2}} + \dfrac{v - u}{1 - \dfrac{uv}{c^2}} = \dfrac{2v\left(1 - \dfrac{u^2}{c^2}\right)}{1 - \dfrac{u^2v^2}{c^4}}$

Since $v < c$, this sum is not equal to $2v$. Therefore, $v_B \neq \dfrac{1}{2}(v_A + v_C)$.

Answer: $v_A > v_B > v_C$