Animated Solution for Physics - Optics: A ray of light travelling in the direction $\frac{1}{2}(\hat{i}+\sqrt{3}\hat{j})$ is incident on a plane mirror. After reflection, it travels along the direction $\frac{1}{2}(\hat{i}-\sqrt{3}\hat{j})$. The angle of incidence is

Imagine you are standing in a room, and a laser beam shoots across the space. Instead of just giving you the angles, what if I gave you the exact mathematical vector of that beam? This problem beautifully marries the geometry of optics with the precision of vectors.

We are given the direction of an incident ray as a vector and its direction after reflecting off a plane mirror. Our mission? To find the angle of incidence. I know vectors can sometimes look intimidating, but let's take a breath and break this down logically.

Let's look closely at the two vectors provided: Incident direction: ${\hat{e}}_i=\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j}$ Reflected direction: ${\hat{e}}_r=\frac{1}{2}\hat{i}-\frac{\sqrt{3}}{2}\hat{j}$

Notice something fascinating? The $x$-component ($\frac{1}{2}\hat{i}$) remains completely unchanged, but the $y$-component has flipped its sign from positive to negative!

What does this physical reality tell us? It means the mirror must be lying horizontally, parallel to the $x$-axis. When the ray hits this horizontal mirror, its horizontal velocity continues unaffected, but its vertical velocity bounces back. Consequently, the normal to the mirror is perfectly vertical, aligning with the $y$-axis.

To find the angle of incidence, we first need to understand the trajectory of the incoming ray. Let's find the angle $\theta$ that the incident ray makes with the horizontal $x$-axis.

For any vector, the tangent of its angle with the horizontal is simply the ratio of its $y$-component to its $x$-component:

Let's substitute our known values into this raw structure:

The $\frac{1}{2}$ in the numerator and denominator elegantly cancel out, leaving us with:

From our standard trigonometric values, we know that the angle whose tangent is $\sqrt{3}$ is $60^{\circ }$. So, the ray strikes the mirror at a $60^{\circ }$ angle to the surface.

There is a catch here! Don't rush and select $60^{\circ }$ as your answer. The angle of incidence is strictly defined as the angle between the incident ray and the normal.

Since the normal is vertical ($90^{\circ }$ to the surface), we simply subtract our glancing angle:

And there we have it! The angle of incidence is $30^{\circ }$.

For those who love pure vector algebra, there is a foolproof alternative. The angle between the incident vector and the reflected vector is always $180^{\circ }-2i$.

By taking the dot product of the two unit vectors:

Since $\cos (120^{\circ })=-\frac{1}{2}$, we get:

Physics is beautifully consistent, no matter which path you choose!