Two plane mirrors M1 and M2 are at right angle to each other shown. A point source P is placed at a

Q: Why do we only consider two images? Aren't there more?

You are absolutely right! Because the mirrors are at $90^\circ$, there are actually $3$ images formed in total (using the formula $n = 360/\theta - 1$). The third image is formed by multiple reflections at $(-a, -2a)$. However, the question specifically asks for the distance between the images formed directly by $M_1$ and $M_2$ as described in the setup.

Q: Can I solve this using geometry instead of coordinates?

Yes, you can! If you draw the setup, you'll see that the distance between $I_1$ and the y-axis is $a$, and the distance from the y-axis to $I_2$ is $a$. The vertical distance is $2a + 2a = 4a$. You can form a right-angled triangle with base $2a$ and height $4a$, and use Pythagoras' theorem to find the hypotenuse, which gives the same result.

Q: What if the angle between the mirrors wasn't 90 degrees?

If the angle wasn't $90^\circ$, the coordinate geometry method becomes much more complex because the reflection lines wouldn't align perfectly with the x and y axes. In such cases, using polar coordinates or drawing circles of reflection is a better approach.

Q: Why is the shortest distance a straight line?

In Euclidean geometry (which applies to our physical space here), the shortest path between any two points is always a straight line. The distance formula calculates exactly the length of this straight line segment.

Q: I got confused with the signs while calculating the distance. How to avoid that?

Always write down the coordinates clearly: $(x_1, y_1)$ and $(x_2, y_2)$. When substituting into $(x_2 - x_1)$, use parentheses for negative numbers, like $(a - (-a))$. This small habit prevents silly sign errors that can ruin the whole calculation.

Question

Two plane mirrors $M_1$ and $M_2$ are at right angle to each other shown. A point source $P$ is placed at $a$ and $2a$ meter away from $M_1$ and $M_2$, respectively. The shortest distance between the images thus formed is (Take $\sqrt{5} = 2.3$)

Options:

(A) $3a$

(B) $4.6a$

(C) $2.3a$

(D) $2\sqrt{10} a$

Sigma Percentile · Accepted Answer

This problem is a beautiful intersection of optics and coordinate geometry. By translating a physical setup into a mathematical framework, we can bypass complex ray diagrams and arrive at the solution with pure algebraic elegance.

### Setting Up the Coordinate System

Imagine the intersection of the two perpendicular mirrors, $M_1$ and $M_2$, as the origin $(0,0)$ of a Cartesian plane. Let the vertical mirror $M_1$ lie along the y-axis and the horizontal mirror $M_2$ lie along the x-axis.

The point source $P$ is placed at a distance $a$ from $M_1$ and $2a$ from $M_2$. In our coordinate system, this translates directly to the coordinates of $P$ being $(a, 2a)$.

### The Magic of Reflection

The fundamental principle of a plane mirror is that the image is formed at the exact same perpendicular distance behind the mirror as the object is in front of it.

When we reflect point $P(a, 2a)$ across mirror $M_1$ (the y-axis), the x-coordinate simply flips its sign. Therefore, the first image, $I_1$, is formed at $(-a, 2a)$.

Similarly, reflecting point $P(a, 2a)$ across mirror $M_2$ (the x-axis) flips the y-coordinate. This gives us the position of the second image, $I_2$, at $(a, -2a)$.

### Calculating the Shortest Distance

The question asks for the shortest distance between these two images, $I_1$ and $I_2$. In a Cartesian plane, the shortest distance between any two points is the straight line connecting them, which we can find using the standard distance formula:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

Substituting the coordinates of $I_1(-a, 2a)$ and $I_2(a, -2a)$ into the formula, we get:

$$d = \sqrt{(a - (-a))^2 + (-2a - 2a)^2}$$

$$d = \sqrt{(2a)^2 + (-4a)^2}$$

$$d = \sqrt{4a^2 + 16a^2}$$

$$d = \sqrt{20a^2}$$

Pulling out the perfect squares, we simplify this to:

$$d = 2\sqrt{5}a$$

### The Final Touch

The problem kindly provides the approximation $\sqrt{5} = 2.3$. Substituting this value into our expression yields the final numerical answer:

$$d = 2 	imes 2.3 	imes a$$

**$$d = 4.6a$$**

This elegant approach not only solves the problem quickly but also builds a strong intuition for handling multiple reflections using coordinate geometry.

Animated Solution for Physics - Optics: Two plane mirrors $M_{1}$ and $M_{2}$ are at right angle to each other shown. A point source $P$ is placed at $a$ and $2 a$ meter away from $M_{1}$ and $M_{2}$ , respectively. The shortest distance between the images thus formed is (Take $5 = 2.3$ )

Select Answer:

Visualized Solution

Coordinate System Setup

Image Formation by Plane Mirror

Image $I_{1}$ due to $M_{1}$

Image $I_{2}$ due to $M_{2}$

Distance Formula Setup

Calculating Distance

Final Substitution

The Way Forward

The Sigma Insight: Plane Mirror

Setting Up the Coordinate System

The Magic of Reflection

Calculating the Shortest Distance

The Final Touch

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