Animated Solution for Physics - Optics: A ray of light is incident normally on one of the faces of a prism of apex angle $30^{\circ }$ and refractive index $\sqrt{2}$. The angle of deviation of the ray is ...... degrees.

Imagine a beam of light traveling through the air, completely undisturbed, until it encounters a beautifully cut glass prism. The way light interacts with such geometric shapes is not just mathematically elegant; it forms the foundation of optics!

In this problem, we are given a prism with an apex angle $A=30^{\circ }$ and a refractive index $\mu =\sqrt{2}$. A ray of light strikes one of its faces, $AB$, perfectly normally. Our mission is to find out how much this ray deviates from its original path by the time it exits the prism.

Face $AB$

When light strikes a boundary between two media, it usually bends. However, there is a special case: normal incidence.

Because our ray hits face $AB$ at exactly $90^{\circ }$ to the surface, its angle of incidence is $0^{\circ }$ relative to the normal. According to Snell's Law, ${\mu }_1\sin (0^{\circ })={\mu }_2\sin (r_1)$, which means the angle of refraction $r_1$ must also be $0^{\circ }$.

The ray marches straight through face $AB$ without any deviation whatsoever! It continues its journey inside the glass until it reaches the second face, $AC$.

Now, things get interesting. The ray is inside the prism, heading towards face $AC$. To figure out what happens next, we need to know its angle of incidence at this second face. Let's call this angle $r_2$.

Let's look at the triangle formed by the apex $A$, the point of entry on $AB$, and the point of exit on $AC$. Since the ray is perpendicular to $AB$, it forms a right-angled triangle with the apex.

The apex angle is $A=30^{\circ }$. Therefore, the angle the ray makes with the face $AC$ is $90^{\circ }-30^{\circ }=60^{\circ }$.

But remember, the angle of incidence is always measured from the normal (the perpendicular line to the surface). Since the normal itself is at $90^{\circ }$ to the face $AC$, the angle of incidence $r_2$ is simply $90^{\circ }-60^{\circ }=30^{\circ }$.

A quicker way to remember this for a ray incident normally on one face is the geometric relation: $r_2=A=30^{\circ }$.

Snell's Law at Face $AC$

The ray is now trying to escape the denser glass ($\mu =\sqrt{2}$) into the rarer air ($\mu =1$). It will bend away from the normal. Let's find the angle of emergence, $i_2$, using Snell's Law:

Substituting our known values:

We know that $\sin (30^{\circ })=\frac{1}{2}$. Let's plug that in:

What angle has a sine of $\frac{1}{\sqrt{2}}$? Exactly $45^{\circ }$!

So, the ray emerges from the prism at an angle $i_2=45^{\circ }$ relative to the normal.

We are almost there! The question asks for the angle of deviation, $\delta$. This is the angle between the ray's original incoming direction and its final outgoing direction.

Since the ray didn't bend at all at the first face, the total deviation is entirely due to the bending at the second face.

At face $AC$, the ray was traveling at $30^{\circ }$ relative to the normal, but it bent to $45^{\circ }$. The deviation is simply the difference between these two angles:

The ray has deviated by $15^{\circ }$ from its original path.

This problem beautifully demonstrates how a solid grasp of basic geometry, combined with Snell's Law, can effortlessly unlock the secrets of optical phenomena!