The plane mirrors (M1 and M2) are inclined to each other such that a ray of light incident on mirror

Q: Why is the angle of the incident ray with the mirror equal to $\theta$?

Because the incident ray is parallel to the second mirror. If you extend the first mirror as a transversal line, the angle between the mirrors and the angle between the incident ray and the first mirror are corresponding angles, which are equal.

Q: Does the law of reflection apply to the angle with the mirror surface?

Yes! While the law of reflection is formally stated using the normal (angle of incidence = angle of reflection), it mathematically implies that the angle the incident ray makes with the mirror surface is exactly equal to the angle the reflected ray makes with the mirror surface.

Q: How did we get $180^\circ - 2\theta$ for the angle at R?

Look at the triangle formed by the intersection of the mirrors (O) and the two points of reflection (Q and R). The angle at O is $\theta$, and the angle at Q is $\theta$. Since the sum of angles in a triangle is $180^\circ$, the third angle at R must be $180^\circ - (\theta + \theta) = 180^\circ - 2\theta$.

Q: Why is the angle of the final reflected ray with the second mirror also $\theta$?

The final reflected ray is parallel to the first mirror. If you treat the second mirror as a transversal line intersecting these two parallel lines, the angle between the mirrors ($\theta$) and the angle between the final ray and the second mirror are corresponding angles.

Q: What if the ray wasn't parallel to the second mirror initially?

If the initial ray had a different angle, say $\alpha$, the entire geometry would shift. The angle at Q would be $\alpha$, the angle at R inside the triangle would be $180^\circ - (\theta + \alpha)$, and the final reflected angle would depend on this new geometry. The specific parallel conditions in this problem are what force $\theta$ to be exactly $60^\circ$.

Question

The plane mirrors ($M_1$ and $M_2$) are inclined to each other such that a ray of light incident on mirror $M_1$ and parallel to the mirror $M_2$ is reflected from mirror $M_2$ parallel to the mirror $M_1$. The angle between the two mirror is

Options:

(A) $45^\circ$

(B) $75^\circ$

(C) $90^\circ$

(D) $60^\circ$

Sigma Percentile · Accepted Answer

### The Dance of Light Between Mirrors

Imagine standing in a room with two giant mirrors, $M_1$ and $M_2$, joined at their base to form an unknown angle $	heta$.

A single beam of light enters this setup, and its path is going to reveal the exact angle between the mirrors. This is a classic problem that beautifully marries the physics of reflection with the elegance of pure geometry.

Let's break down the journey of this light ray step by step.

### The First Strike: Parallel Precision

The problem states that the initial light ray, let's call it $PQ$, comes in perfectly parallel to the second mirror, $M_2$. It travels and strikes the first mirror, $M_1$, at a point $Q$.

Because the ray $PQ$ is parallel to $M_2$, we can use a fundamental rule of geometry: **corresponding angles are equal**. If we treat $M_1$ as a transversal line intersecting the parallel lines $PQ$ and $M_2$, the angle the incident ray makes with $M_1$ is exactly equal to the angle between the mirrors, $	heta$.

### The Law of Reflection Takes Over

Now, the light ray hits $M_1$ and bounces off. According to the **Law of Reflection**, the angle of incidence equals the angle of reflection.

While we usually measure these angles from the normal (the perpendicular line), this law also implies that the angle the ray makes with the *surface* of the mirror remains the same after reflection. Therefore, the reflected ray, $QR$, also makes an angle $	heta$ with the mirror $M_1$.

### The Geometric Triangle

Let's pause and look at the shape formed by the origin $O$ (where the mirrors meet) and the two points of reflection, $Q$ and $R$. This forms a triangle, $	riangle OQR$.

We already know two angles inside this triangle: the angle at $O$ is $	heta$, and the angle at $Q$ is also $	heta$. Since the sum of all angles in any triangle is always $180^\circ$, we can easily find the third angle at $R$.

The angle $\angle ORQ$ must be $180^\circ - (	heta + 	heta)$, which simplifies to $180^\circ - 2	heta$.

### The Final Bounce

The ray $QR$ travels to the second mirror, $M_2$, and reflects off it as a new ray, $RS$. The problem gives us a crucial piece of information here: this final reflected ray $RS$ is perfectly parallel to the first mirror, $M_1$.

Because $RS$ is parallel to $M_1$, we can again use the property of corresponding angles. The angle that $RS$ makes with the transversal mirror $M_2$ must be equal to the angle between the mirrors, $	heta$.

### Bringing It All Together

We apply the **Law of Reflection** one last time at mirror $M_2$. The angle the reflected ray $RS$ makes with the mirror ($	heta$) must be exactly equal to the angle the incident ray $QR$ makes with the mirror.

This means that the angle $\angle ORQ$ is also equal to $	heta$.

But wait! Earlier, we found that $\angle ORQ$ is equal to $180^\circ - 2	heta$ from our triangle geometry. This gives us a beautiful equation:

$$180^\circ - 2	heta = 	heta$$

By adding $2	heta$ to both sides, we get:

$$3	heta = 180^\circ$$

Dividing by 3, we arrive at our final answer:

$$	heta = 60^\circ$$

The mirrors are inclined at exactly $60^\circ$. The symmetry of the light's path perfectly locks the geometry into this single, elegant solution!

Animated Solution for Physics - Optics: The plane mirrors ( $M_{1}$ and $M_{2}$ ) are inclined to each other such that a ray of light incident on mirror $M_{1}$ and parallel to the mirror $M_{2}$ is reflected from mirror $M_{2}$ parallel to the mirror $M_{1}$ . The angle between the two mirror is

Select Answer:

Visualized Solution

Angle Between Mirrors $M_{1}$ and $M_{2}$

Incident Ray $P Q ∥ M_{2}$

Reflection at Mirror $M_{1}$

Angles in $△ O QR$

Reflected Ray $R S ∥ M_{1}$

Corresponding Angle at $M_{2}$

Reflection at Mirror $M_{2}$

Equating $∠ O R Q$

Solving for $θ$

The Sigma Insight: Plane Mirror

The Dance of Light Between Mirrors

The First Strike

The Law of Reflection Takes Over

The Geometric Triangle

The Final Bounce

Bringing It All Together

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Visualized Solution

Angle Between Mirrors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M2​

Incident Ray .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ∥M2​

Reflection at Mirror .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M1​

Angles in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △OQR

Reflected Ray .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } RS∥M1​

Corresponding Angle at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M2​

Reflection at Mirror .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M2​

Equating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∠ORQ

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ