A diver looking up through the water sees the outside world contained in a circular horizon. The refractive index of water is $$\frac{4}{3}$$, and the diver's eyes are 15 cm below the surface of the water. Then the radius of the circle is:

The diver is looking up from below the water surface, and the circular horizon is formed due to the phenomenon of total internal reflection. The refractive index of water, denoted as $$\mu$$, is given as $$\frac{4}{3}$$. The diver's eyes are at a depth of 15 cm below the water surface.

To find the radius of the circular horizon, we consider the critical angle, $$\theta_c$$, which is the angle of incidence in the denser medium (water) for which the angle of refraction in air is 90 degrees. The formula for the critical angle is:

$$\sin \theta_c = \frac{1}{\mu}$$

Substituting $$\mu = \frac{4}{3}$$:

$$\sin \theta_c = \frac{1}{\frac{4}{3}} = \frac{3}{4}$$

So, $$\sin \theta_c = \frac{3}{4}$$.

Now, light rays from the horizon are very far away and travel horizontally in air. When such a ray hits the water surface at point A, the angle of incidence in air is 90 degrees because the ray is horizontal and the normal to the water surface is vertical. Applying Snell's law at the air-water interface:

$$\mu_{\text{air}} \sin i = \mu_{\text{water}} \sin r$$

Since $$\mu_{\text{air}} \approx 1$$ and $$i = 90^\circ$$:

$$1 \cdot \sin 90^\circ = \frac{4}{3} \cdot \sin r$$

$$\sin 90^\circ = 1$$, so:

$$1 = \frac{4}{3} \sin r$$

Solving for $$\sin r$$:

$$\sin r = \frac{3}{4}$$

Thus, the angle of refraction in water, $$r$$, equals the critical angle $$\theta_c$$, so $$r = \theta_c$$ and $$\sin r = \frac{3}{4}$$.

In the water, the refracted ray makes an angle $$r$$ with the normal (vertical). Therefore, the angle between the refracted ray and the horizontal is $$90^\circ - r$$. The diver's eye is at point E, 15 cm directly below point O on the water surface. The refracted ray travels from point A on the water surface to point E. Let the horizontal distance OA be the radius $$x$$ cm, which we need to find.

Consider the right triangle OAE, where:

• O is the point directly above E on the water surface,
• A is the point where the refracted ray meets the water surface,
• E is the diver's eye.

This triangle is right-angled at O, with OE = 15 cm (vertical) and OA = $$x$$ cm (horizontal). The angle at A in triangle OAE is the angle between OA (horizontal) and AE (the refracted ray), which is $$90^\circ - r$$.

In triangle OAE:

• Opposite side to angle at A is OE = 15 cm,
• Adjacent side to angle at A is OA = $$x$$ cm.

Therefore:

$$\tan(90^\circ - r) = \frac{\text{opposite}}{\text{adjacent}} = \frac{15}{x}$$

Since $$\tan(90^\circ - r) = \cot r$$:

$$\cot r = \frac{15}{x}$$

Solving for $$x$$:

$$x = 15 \cdot \tan r$$

We know $$\sin r = \frac{3}{4}$$. To find $$\tan r$$, we use:

$$\tan r = \frac{\sin r}{\cos r}$$

First, find $$\cos r$$:

$$\cos r = \sqrt{1 - \sin^2 r} = \sqrt{1 - \left(\frac{3}{4}\right)^2} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$$

Now,

$$\tan r = \frac{\sin r}{\cos r} = \frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}} = \frac{3}{4} \cdot \frac{4}{\sqrt{7}} = \frac{3}{\sqrt{7}}$$

Substitute into the expression for $$x$$:

$$x = 15 \cdot \frac{3}{\sqrt{7}} = \frac{45}{\sqrt{7}} \text{ cm}$$

This matches option B: $$\frac{15 \times 3}{\sqrt{7}} \text{ cm}$$.

Hence, the correct answer is Option B.