Animated Solution for Physics - Optics: A ray of light from a denser medium strikes a rarer medium at an angle of incidence $i$ (see figure). The reflected and refracted rays make an angle of $90^{\circ }$ with each other. The angles of reflection and refraction are $r$ and $r^{\prime }$. The critical angle is

This problem is a beautiful interplay of basic geometry, the law of reflection, and Snell's Law. Let's break down the thought process step-by-step to see how these fundamental principles lead us straight to the answer.

Imagine a ray of light traveling through a denser medium and striking the boundary of a rarer medium. At this interface, two things happen simultaneously: part of the light is reflected back into the denser medium, and part of it is refracted into the rarer medium.

We are given a crucial piece of geometric information: the angle between the reflected ray and the refracted ray is exactly $90^{\circ }$.

If we look at the normal (the imaginary perpendicular line at the point of incidence), the angles on one side of it must add up to a straight line, which is $180^{\circ }$. These angles are the angle of reflection $r$, the $90^{\circ }$ gap between the rays, and the angle of refraction $r^{\prime }$.

Therefore, we can write:

By simply rearranging this, we can express the angle of refraction in terms of the angle of reflection:

Snell's Law

Now, let's bring in the physics. According to the law of reflection, the angle of incidence $i$ is always equal to the angle of reflection $r$. So, we know that $i=r$.

Next, we apply Snell's Law at the interface, which relates the angles to the refractive indices of the two media:

Let's substitute the relationships we just found ($i=r$ and $r^{\prime }=90^{\circ }-r$) into Snell's Law:

From basic trigonometry, we know that $\sin (90^{\circ }-r)$ is simply $\cos r$. Substituting this identity, our equation becomes much cleaner:

Let's group the trigonometric terms on one side and the refractive indices on the other. Dividing both sides by $\cos r$ and ${\mu }_D$, we get:

Which simplifies to:

Now, the question asks for the critical angle, ${\theta }_C$. The critical angle is defined as the angle of incidence for which the angle of refraction is exactly $90^{\circ }$. Mathematically, it is given by the sine inverse of the ratio of the rarer medium's refractive index to the denser medium's refractive index:

Since we just proved that $\frac{{\mu }_R}{{\mu }_D}=\tan r$, we can substitute this directly into our critical angle formula:

Taking the sine inverse of both sides gives us our final answer:

$${\theta }_C={\sin }^{-1}(\tan r)$$

This perfectly matches the first option. By systematically applying geometry and fundamental optical laws, a seemingly complex problem unravels into an elegant solution.