Sound can travel from point A to point B via two main types of paths:

1. Direct Path: Directly through the water.
2. Refracted Path: Along the rod for some distance \(x\), and then refracting through the water at an angle to reach B.

Let the horizontal separation between A and B be \(l\). The vertical separation is \(d\).

Let the sound travel a distance \(x\) along the rod with speed \(v\), and then travel through water with speed \(c\). The remaining horizontal distance to cover in water is \(l-x\). The distance through water is \(\sqrt{(l-x)^2 + d^2}\). The total time function \(t(x)\) is: \[ t(x) = \frac{x}{v} + \frac{\sqrt{(l-x)^2 + d^2}}{c} \] To find the minimum time, we differentiate with respect to \(x\) and set to zero: \[ \frac{dt}{dx} = \frac{1}{v} + \frac{1}{c} \frac{1}{2\sqrt{(l-x)^2 + d^2}} \cdot 2(l-x)(-1) = 0 \] \[ \frac{1}{v} = \frac{l-x}{c\sqrt{(l-x)^2 + d^2}} \]

Let \(\theta\) be the angle the water path makes with the rod. Then \(\cos \theta = \frac{l-x}{\sqrt{(l-x)^2 + d^2}}\). The optimization condition becomes: \[ \frac{1}{v} = \frac{\cos \theta}{c} \implies \cos \theta = \frac{c}{v} \] This is the condition for the “head wave” or critical refraction. Using \(\sin \theta = \sqrt{1 – \cos^2 \theta} = \sqrt{1 – \frac{c^2}{v^2}}\): \[ t_{min} = \frac{l – d \cot \theta}{v} + \frac{d}{c \sin \theta} = \frac{l}{v} + d \sqrt{\frac{1}{c^2} – \frac{1}{v^2}} \] This solution is valid only if \(x > 0\), which requires the rod to be fast enough such that the optimal angle hits the second rod within the length \(l\). Condition: \(\frac{c}{v} < \frac{l}{\sqrt{l^2 + d^2}}\).

Case 1: If \(\frac{c}{v} \ge \frac{l}{\sqrt{l^2 + d^2}}\), the optimal path is the direct line connecting A and B (corner solution at \(x=0\)). \[ t = \frac{\sqrt{d^2 + l^2}}{c} \] Case 2: If \(\frac{c}{v} < \frac{l}{\sqrt{l^2 + d^2}}\), the refracted path is faster. \[ t = \frac{l}{v} + d \sqrt{\frac{1}{c^2} - \frac{1}{v^2}} \]