1. Geometric Setup and Ray Diagram

The visible area of the seabed is determined by the maximum angle at which light can enter the submarine window and still emerge into the air inside. This limiting condition corresponds to the ray grazing the inner surface of the window (angle of emergence $\approx 90^\circ$).

Parameters:

• Window radius $r = 30 \text{ cm} = 0.3 \text{ m}$
• Depth of seabed $d = 6.0 \text{ m}$
• Refractive index of water $\mu_w = 1.25$
• Refractive index of glass $\mu_g = 1.5$ (Note: Since $\mu_g > \mu_w$, the critical limit is determined by the Water-Air effective interface).

2. Derivation

Let the critical angle in water be $C$. This is the angle incident on the window that results in grazing emergence ($90^\circ$) into the air inside the submarine.

Using Snell’s Law:

$$ \mu_w \sin C = \mu_{air} \sin 90^\circ $$ $$ 1.25 \sin C = 1 \cdot 1 $$ $$ \sin C = \frac{1}{1.25} = \frac{4}{5} $$

We need $\tan C$ to calculate the horizontal distance. Using the trigonometric identity:

$$ \tan C = \frac{\sin C}{\sqrt{1 – \sin^2 C}} = \frac{4/5}{\sqrt{1 – (4/5)^2}} = \frac{4/5}{\sqrt{9/25}} = \frac{4/5}{3/5} = \frac{4}{3} $$

The total radius $R$ of the visible portion of the seabed is the sum of the window’s radius $r$ and the horizontal expansion $x$ due to the depth $d$:

$$ R = r + x $$ $$ x = d \tan C $$ $$ R = r + d \tan C $$

3. Calculation

Substituting the values:

$$ R = 0.30 \text{ m} + 6.0 \text{ m} \times \frac{4}{3} $$ $$ R = 0.30 + 2.0 \times 4 $$ $$ R = 0.30 + 8.0 = 8.3 \text{ m} $$