A

Step 1: Understand the Critical Angle.
The critical angle (B1) is the angle of incidence above which total internal reflection occurs. It is defined as:
$$ n = \frac{1}{\sin(\alpha)} $$, where n is the refractive index of the medium (in this case, glass).

Step 2: Analyze the Incident Ray.
A horizontal beam of light enters face AB at grazing incidence, meaning the angle of incidence (i) is 90 degrees. At this point, using Snell's law:
$$ n_{air} \cdot \sin(90^\circ) = n_{glass} \cdot \sin(\alpha) $$
This implies that since n_{air} is 1, we have:
$$ 1 = n_{glass} \cdot \sin(\alpha) \Rightarrow n_{glass} = \frac{1}{\sin(\alpha)} $$

Step 3: Angle of Refraction.
According to Snell's law, at the interface between the glass and the air, when light emerges from face BC, let θ be the angle which the ray makes with the normal to BC. Applying Snell's Law again, we have:
$$ n_{glass} \cdot \sin(\theta) = n_{air} \cdot \sin(90-\alpha) $$
Since n_{air} = 1, this simplifies to:
$$ n_{glass} \cdot \sin(\theta) = \cos(\alpha) $$
Substituting the value of n_{glass} from Step 2, we get:
$$ \frac{1}{\sin(\alpha)} \cdot \sin(\theta) = \cos(\alpha) $$
Rearranging gives:
$$ \sin(\theta) = \sin(\alpha) \cdot \cos(\alpha) $$

Step 4: Using Trigonometric Identity.
We know that:
$$ \sin(\alpha) \cdot \cos(\alpha) = \frac{\sin(2\alpha)}{2} $$
So using cotangent related identities:
$$ \sin(\theta) = \cot(\alpha) $$
Thus proving that sin θ = cot α.

Step 5: Greatest Refractive Index.
For light to emerge from face BC, we must have:
$$ n_{glass} \cdot \sin(\alpha) \leq 1 $$
Hence, substituting the inequality we derived:
$$ n_{glass} \leq \frac{1}{\sin(\alpha)} \Rightarrow \text{Greatest value of } n_{glass} = 1 $$ when B1 approaches 90 degrees, or when total internal reflection is about to occur.