A ray of light coming from the point (2, 2√3) is incident at an angle 30° on the line x = 1 at the p

Q: Wait, why did you use the angle $60^\circ$ for the slope instead of $30^\circ$?

Great catch! In optics, the angle of incidence is defined relative to the normal (the line perpendicular to the mirror). Since the mirror is the vertical line $x=1$, the normal is a horizontal line. If the ray makes a $30^\circ$ angle with the mirror, it must make a $90^\circ - 30^\circ = 60^\circ$ angle with the normal. Always draw your normal line first to avoid this trap!

Q: Why can't I just use the law of reflection with angles directly?

You absolutely can! However, using the virtual image method is a powerful 'JEE-pro' shortcut. It transforms a reflection problem into a simple coordinate geometry problem involving a straight line. It saves time and reduces the chance of sign errors in trigonometry.

Q: How do I know the virtual image $P'$ is at $(0, 2\sqrt{3})$?

The mirror is the line $x=1$. The point $P$ is at $x=2$. Since $P$ is $1$ unit to the right of the mirror, its reflection $P'$ must be $1$ unit to the left of the mirror. Moving $1$ unit left from $x=1$ lands you at $x=0$. The $y$-coordinate remains unchanged because the mirror is vertical.

Q: Is it possible for the reflected ray to have a positive slope?

Think about the geometry: the ray hits the mirror and bounces 'downward' toward the $x$-axis. Since it is moving from a higher $y$ to a lower $y$ as $x$ increases, the slope must be negative. If you ever get a positive slope in a reflection problem like this, check your coordinate signs!

Q: Do I really need to find the point $B$ on the $x$-axis?

Actually, no! The problem asks which point lies on the line $AB$. Once you have the equation of the line $AB$, you can simply test the given options. If a point satisfies the equation, it lies on the line. It's a classic time-saving strategy in competitive exams.

Question

A ray of light coming from the point $(2, 2\sqrt{3})$ is incident at an angle $30^\circ$ on the line $x = 1$ at the point $A$. The ray gets reflected on the line $x = 1$ and meets $x$-axis at the point $B$. Then, the line $AB$ passes through the point:

Options:

(A) $(3, -\frac{1}{\sqrt{3}})$

(B) $(3, -\sqrt{3})$

(C) $(4, -\sqrt{3})$

(D) $(4, -\frac{\sqrt{3}}{2})$

Sigma Percentile · Accepted Answer

### The Dance of Light: A Journey Through Reflection

Imagine you are standing in a dark room, holding a laser pointer. You aim it at a mirror, watching the beam dance across the wall. In the world of JEE Advanced, this isn't just a light show; it is a beautiful interplay of geometry and algebra.

Today, we are going to master the reflection of a ray of light off a vertical mirror, turning a seemingly complex path into a simple, elegant line equation.

### Phase 1: The Incident Ray

We begin with a source point $P(2, 2\sqrt{3})$ and a mirror defined by the line $x = 1$. The ray strikes the mirror at point $A$.

The problem states the ray makes an angle of $30^\circ$ with the mirror. Since our mirror is the vertical line $x=1$, the normal is a horizontal line. Thus, the angle the ray makes with the normal is $90^\circ - 30^\circ = 60^\circ$.

To find the slope of the incident ray $PA$, we use the tangent of the angle it makes with the horizontal: $m = 	an(60^\circ) = \sqrt{3}$. Since the ray passes through $P(2, 2\sqrt{3})$ and $A$ lies on $x=1$, we use the slope formula:

$$
\frac{2\sqrt{3} - y_A}{2 - 1} = \sqrt{3}
$$

Solving this, we find $2\sqrt{3} - y_A = \sqrt{3}$, which gives us $y_A = \sqrt{3}$. So, our point of incidence is $A(1, \sqrt{3})$.

### Phase 2: The Virtual Image Shortcut

We could use the law of reflection to find the angle of the reflected ray, but there is a more elegant way. In physics, we know that a reflected ray appears to originate from the virtual image of the source point.

Let $P'$ be the reflection of $P(2, 2\sqrt{3})$ across the line $x=1$. Since $P$ is $1$ unit to the right of the mirror, $P'$ must be $1$ unit to the left. Thus, $P' = (0, 2\sqrt{3})$.

This is the 'Aha!' moment. The reflected ray is simply the straight line connecting the virtual image $P'$ and the point of incidence $A$. We have effectively turned a reflection problem into a simple line-through-two-points problem.

### Phase 3: The Final Equation

We now have two points on our reflected ray: $P'(0, 2\sqrt{3})$ and $A(1, \sqrt{3})$. The slope $m'$ of this line is:

$$
m' = \frac{\sqrt{3} - 2\sqrt{3}}{1 - 0} = -\sqrt{3}
$$

Using the point-slope form $y - y_1 = m(x - x_1)$ with point $P'$, we get:

$$
y - 2\sqrt{3} = -\sqrt{3}(x - 0)
$$

This simplifies beautifully to the final equation of the reflected ray:

$$
y = -\sqrt{3}x + 2\sqrt{3}
$$

### Phase 4: The Victory Lap

We are looking for a point that lies on this line. We test the provided coordinate $(3, -\sqrt{3})$:

$$
y = -\sqrt{3}(3) + 2\sqrt{3} = -3\sqrt{3} + 2\sqrt{3} = -\sqrt{3}
$$

It matches perfectly! The point **$(3, -\sqrt{3})$** lies exactly on the path of the reflected ray.

You have successfully navigated the geometry, utilized the power of virtual images, and arrived at the solution with mathematical precision. Remember, in physics, every complex path is just a collection of simple, logical steps.

Select Answer:

Visualized Solution

Setting up the Coordinate System

The Incident Ray and Angle

Slope of the Incident Ray

Setting up Coordinates for Point $A$

Calculating the Coordinates of $A$

The Reflection Principle

Coordinates of Image Point $P^{'}$

Tracing the Reflected Ray

Slope of the Reflected Ray

Calculating the Slope $m^{'}$

Equation of the Reflected Ray

Checking the Given Options

Final Confirmation

The Sigma Insight: Various Forms of Equations of a Line

The Dance of Light

Phase 1

Phase 2

Phase 3

Phase 4

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