We need to find the minimum time for sound to travel between two ships anchored at a separation $l$. The sound can propagate through three different media:

• Air: Direct path above surface ($v_a = 333 \text{ m/s}$).
• Water: Direct path through the water ($v_w = 1.5 \text{ km/s}$).
• Rock (Seabed): A path that involves refraction: traveling down to the seabed, moving along the rock interface, and traveling back up ($v_r = 4.5 \text{ km/s}$).

Since the speed of sound in rock ($v_r$) is significantly higher than in water ($v_w$) or air ($v_a$), the path involving the rock layer might be the fastest, despite the extra distance traveled vertically. This is a classic example of a “head wave” or lateral wave propagation, governed by Fermat’s Principle of Least Time.

Let $t$ be the total time taken for the sound to travel from ship A to ship B via the critical path shown above. The path consists of traveling distance $h/\cos\theta$ in water twice, and traveling the remaining horizontal distance in the rock.

The total horizontal distance is $l$. The horizontal distance covered in water is $2h \tan\theta$. Therefore, the distance traveled in rock is $l – 2h \tan\theta$.

The total time equation is:

$$ t = \frac{2h}{v_w \cos \theta} + \frac{l – 2h \tan \theta}{v_r} $$

For the wave to travel along the interface, the angle of incidence $\theta$ must be the critical angle. By Snell’s Law:

$$ \sin \theta = \frac{v_w}{v_r} $$

Substituting $\frac{1}{v_r} = \frac{\sin \theta}{v_w}$ into the time equation:

$$ \begin{aligned} t &= \frac{2h}{v_w \cos \theta} + \frac{l}{v_r} – \frac{2h \tan \theta}{v_r} \\ t &= \frac{l}{v_r} + \frac{2h}{v_w \cos \theta} – \frac{2h}{v_w} \cdot \frac{\sin \theta}{\cos \theta} \cdot \sin \theta \quad \left(\text{using } \frac{1}{v_r} = \frac{\sin \theta}{v_w}\right) \\ t &= \frac{l}{v_r} + \frac{2h}{v_w \cos \theta} (1 – \sin^2 \theta) \\ t &= \frac{l}{v_r} + \frac{2h}{v_w \cos \theta} (\cos^2 \theta) \\ t &= \frac{l}{v_r} + \frac{2h \cos \theta}{v_w} \end{aligned} $$

Now, we express $\cos \theta$ in terms of velocities using $\sin \theta = \frac{v_w}{v_r}$:

$$ \cos \theta = \sqrt{1 – \sin^2 \theta} = \sqrt{1 – \frac{v_w^2}{v_r^2}} = \frac{\sqrt{v_r^2 – v_w^2}}{v_r} $$

Substituting this back into the expression for $t$:

$$ t = \frac{l}{v_r} + \frac{2h}{v_w} \left( \frac{\sqrt{v_r^2 – v_w^2}}{v_r} \right) $$ $$ t = \frac{l}{v_r} + \frac{2h\sqrt{v_r^2 – v_w^2}}{v_w v_r} $$

We are given:

• $l = 3 \text{ km}$
• $h = 1 \text{ km}$
• $v_w = 1.5 \text{ km/s}$
• $v_r = 4.5 \text{ km/s}$
• (Note: We perform calculations in km and km/s to keep units consistent).

Substituting the values into the derived formula:

$$ t = \frac{3}{4.5} + \frac{2(1)\sqrt{(4.5)^2 – (1.5)^2}}{1.5 \times 4.5} $$

Simplifying the terms:

$$ t = \frac{2}{3} + \frac{2\sqrt{20.25 – 2.25}}{6.75} $$ $$ t \approx 0.667 + \frac{2\sqrt{18}}{6.75} $$ $$ t \approx 0.667 + \frac{2(4.2426)}{6.75} $$ $$ t \approx 0.667 + \frac{8.4852}{6.75} $$ $$ t \approx 0.667 + 1.257 $$ $$ t \approx 1.924 \text{ s} $$