Let the refractive index of a denser medium with respect to a rarer medium be $$n_{12}$$ and its critical angle be $$\theta_C$$. At an angle of incidence $$A$$ when light is travelling from denser medium to rarer medium, a part of the light is reflected and the rest is refracted and the angle between reflected and refracted rays is 90°. Angle $$A$$ is given by:

We have light travelling from a denser medium (refractive index $$n_1$$) to a rarer medium (refractive index $$n_2$$) with $$n_1 > n_2$$. By definition, the refractive index of the denser medium with respect to the rarer one is

$$ n_{12} \;=\; \frac{n_1}{n_2}. $$

The critical angle $$\theta_C$$ is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium becomes $$90^\circ$$. The well-known relation for the critical angle is stated first:

$$ \sin\theta_C \;=\; \frac{n_2}{n_1}. $$

Because $$n_{12} = \dfrac{n_1}{n_2}$$, the above expression can be rewritten as

$$ \sin\theta_C \;=\; \frac{1}{n_{12}}. \quad -(1) $$

Now let the light be incident at an angle $$A$$ on the interface. A part is reflected and a part is refracted, and the angle between the reflected and refracted rays is given to be $$90^\circ$$.

According to the law of reflection, the angle of reflection equals the angle of incidence, hence the reflected ray makes an angle $$A$$ with the normal. Let the angle of refraction be $$r$$. The reflected ray lies in the denser medium on one side of the normal, while the refracted ray lies in the rarer medium on the other side. Therefore the angle between these two rays is simply the sum of the two individual angles with the normal:

$$ A + r = 90^\circ. \quad -(2) $$

From Snell’s law, which we state before using,

$$ n_1 \sin A = n_2 \sin r, $$

dividing both sides by $$n_2$$ gives

$$ \frac{n_1}{n_2}\,\sin A = \sin r, $$

and since $$\dfrac{n_1}{n_2} = n_{12}$$, we have

$$ n_{12}\,\sin A = \sin r. \quad -(3) $$

Using relation (2), $$r = 90^\circ - A$$, so

$$ \sin r = \sin(90^\circ - A) = \cos A. $$

Substituting this value of $$\sin r$$ in equation (3), we get

$$ n_{12}\,\sin A = \cos A. $$

Dividing both sides by $$\cos A$$ and by $$n_{12}$$ in turn,

$$ \tan A = \frac{1}{n_{12}}. \quad -(4) $$

But from equation (1), $$\dfrac{1}{n_{12}} = \sin\theta_C$$, hence equation (4) becomes

$$ \tan A = \sin\theta_C. $$

Taking the inverse tangent (arctangent) of both sides, we finally obtain

$$ A = \tan^{-1}\!\bigl(\sin\theta_C\bigr). $$

Hence, the correct answer is Option A.