Animated Solution for Physics - Optics: A ray of light undergoes deviation of $30^{\circ }$ when incident on an equilateral prism of refractive index $\sqrt{2}$. The angle made by the ray inside the prism with the base of the prism is ______

The journey through optics often brings us face-to-face with the elegant geometry of prisms. In this problem, we are tasked with finding the angle a refracted ray makes with the base of an equilateral prism. Let's dive into the physics and uncover the magic of minimum deviation!

Imagine a ray of light striking an equilateral prism. Because it is equilateral, we immediately know the prism angle is $A=60^{\circ }$. The problem also provides the refractive index of the prism, $\mu =\sqrt{2}$, and tells us that the ray undergoes a deviation of $\delta =30^{\circ }$.

Our goal is to determine the path of the ray inside the prism, specifically the angle it makes with the base. To do this, we need to check if there is anything special about this $30^{\circ }$ deviation.

In prism optics, there is a beautiful relationship between the refractive index, the prism angle, and the angle of minimum deviation (${\delta }_m$). The formula is:

This equation is our golden key. Let's calculate the theoretical minimum deviation for our specific prism and see what we find.

We plug in our known values: $\mu =\sqrt{2}$ and $A=60^{\circ }$.

Simplifying the denominator, we get $\sin (30^{\circ })$, which we know is $\frac{1}{2}$.

Now, let's isolate the sine term. Multiplying both sides by $\frac{1}{2}$:

We know from trigonometry that $\sin (45^{\circ })=\frac{1}{\sqrt{2}}$. Therefore, the angle inside the sine function must be $45^{\circ }$.

Look at that! Our calculated theoretical minimum deviation is exactly $30^{\circ }$. And what was the deviation given in the problem? Also $30^{\circ }$!

This is a crucial realization: The ray in this problem is undergoing minimum deviation.

Why does this matter? Because at the state of minimum deviation, a prism exhibits perfect optical symmetry. The angle of incidence equals the angle of emergence ($i=e$), and the angle of refraction at the first face equals the angle of incidence at the second face ($r_1=r_2$).

For an isosceles or equilateral prism, this symmetry dictates that the refracted ray traveling inside the prism must be perfectly parallel to the base of the prism.

Since the ray inside the prism is parallel to the base, the angle it makes with the base is simply $0^{\circ }$. The geometry perfectly aligns, giving us a clean and elegant solution!