A ray of light is incident from a denser to a rarer medium. The critical angle for total internal reflection is $$\theta_{iC}$$ and Brewster's angle of incidence is $$\theta_{iB}$$, such that $$\frac{\sin\theta_{iC}}{\sin\theta_{iB}} = \eta = 1.28$$. The relative refractive index of the two media is:

We are given that a ray of light is incident from a denser medium to a rarer medium. The critical angle for total internal reflection is denoted by $$\theta_{iC}$$ and Brewster's angle of incidence is $$\theta_{iB}$$. The ratio $$\frac{\sin\theta_{iC}}{\sin\theta_{iB}} = \eta = 1.28$$ is provided. We need to find the relative refractive index, which is the ratio of the refractive index of the rarer medium to that of the denser medium, denoted as $$\mu = \frac{\mu_r}{\mu_d}$$, where $$\mu_d$$ is the refractive index of the denser medium and $$\mu_r$$ is that of the rarer medium.

First, recall the formula for the critical angle $$\theta_{iC}$$. When light travels from a denser medium to a rarer medium, total internal reflection occurs at the critical angle, given by:

$$\sin\theta_{iC} = \frac{\mu_r}{\mu_d} = \mu$$

So, we have:

$$\sin\theta_{iC} = \mu \quad \text{(Equation 1)}$$

Next, Brewster's angle $$\theta_{iB}$$ is the angle of incidence at which the reflected light is completely polarized. For light going from denser to rarer medium, Brewster's angle is given by:

$$\tan\theta_{iB} = \frac{\mu_r}{\mu_d} = \mu$$

So, we have:

$$\tan\theta_{iB} = \mu \quad \text{(Equation 2)}$$

The given ratio is $$\frac{\sin\theta_{iC}}{\sin\theta_{iB}} = \eta = 1.28$$. Substituting from Equation 1, $$\sin\theta_{iC} = \mu$$, so:

$$\frac{\mu}{\sin\theta_{iB}} = 1.28 \quad \text{(Equation 3)}$$

From Equation 2, $$\tan\theta_{iB} = \mu$$. We need to express $$\sin\theta_{iB}$$ in terms of $$\mu$$. Using the trigonometric identity $$\tan\theta = \frac{\sin\theta}{\cos\theta}$$ and $$\sin^2\theta + \cos^2\theta = 1$$, we can write:

Let $$\theta = \theta_{iB}$$, so $$\tan\theta = \mu$$. Then:

$$\frac{\sin\theta}{\cos\theta} = \mu$$

Therefore, $$\sin\theta = \mu \cos\theta$$. Squaring both sides:

$$\sin^2\theta = \mu^2 \cos^2\theta$$

Substituting $$\sin^2\theta = 1 - \cos^2\theta$$:

$$1 - \cos^2\theta = \mu^2 \cos^2\theta$$

Rearranging terms:

$$1 = \cos^2\theta + \mu^2 \cos^2\theta$$

$$1 = \cos^2\theta (1 + \mu^2)$$

Solving for $$\cos^2\theta$$:

$$\cos^2\theta = \frac{1}{1 + \mu^2}$$

Now, $$\sin^2\theta = 1 - \cos^2\theta = 1 - \frac{1}{1 + \mu^2} = \frac{1 + \mu^2 - 1}{1 + \mu^2} = \frac{\mu^2}{1 + \mu^2}$$. Taking the positive square root (since $$\theta_{iB}$$ is an acute angle):

$$\sin\theta_{iB} = \frac{\mu}{\sqrt{1 + \mu^2}} \quad \text{(Equation 4)}$$

Substituting Equation 4 into Equation 3:

$$\frac{\mu}{\frac{\mu}{\sqrt{1 + \mu^2}}} = 1.28$$

Simplifying the expression:

$$\mu \times \frac{\sqrt{1 + \mu^2}}{\mu} = \sqrt{1 + \mu^2} = 1.28$$

So, we have:

$$\sqrt{1 + \mu^2} = 1.28$$

Squaring both sides to eliminate the square root:

$$1 + \mu^2 = (1.28)^2$$

Calculating $$(1.28)^2$$:

$$1.28 \times 1.28 = 1.6384$$

So:

$$1 + \mu^2 = 1.6384$$

Subtracting 1 from both sides:

$$\mu^2 = 1.6384 - 1 = 0.6384$$

Taking the square root:

$$\mu = \sqrt{0.6384}$$

We know that $$0.8^2 = 0.64$$, which is close to 0.6384. Calculating more precisely, $$0.799^2 = (0.8 - 0.001)^2 = 0.8^2 - 2 \times 0.8 \times 0.001 + (0.001)^2 = 0.64 - 0.0016 + 0.000001 = 0.638401$$, which is approximately 0.6384. Thus, $$\mu \approx 0.799$$, which rounds to 0.8. Verifying with $$\mu = 0.8$$:

$$\sqrt{1 + (0.8)^2} = \sqrt{1 + 0.64} = \sqrt{1.64} \approx \sqrt{1.64}$$

Since $$\sqrt{1.64} = \sqrt{\frac{164}{100}} = \frac{\sqrt{164}}{10} = \frac{2\sqrt{41}}{10} = \frac{\sqrt{41}}{5}$$. Approximating $$\sqrt{41} \approx 6.403$$, so $$\frac{6.403}{5} = 1.2806$$, which is approximately 1.28. Thus, $$\mu = 0.8$$ satisfies the equation.

Comparing with the options:

A. 0.4

B. 0.2

C. 0.9

D. 0.8

The relative refractive index is 0.8, which corresponds to Option D. The problem states that the correct answer is Option 4, and since D is the fourth option, it matches.

Hence, the correct answer is Option D.