Determining the Angle of Incidence for Light Rays Between Two Mirrors

This problem explores the reflection of light between two plane mirrors inclined at a specific angle. We are given that an incident ray on the first mirror ($M_1$) is parallel to the second mirror ($M_2$), and the ray finally reflected from $M_2$ is parallel to $M_1$. Our goal is to find the angle of incidence on $M_1$. Let's break down the physics and geometry involved.

Problem Analysis and Definitions

Here's a breakdown of the setup and the key terms:

• Mirrors: We have two plane mirrors, labeled $M_1$ and $M_2$.
• Mirror Angle: The angle between $M_1$ and $M_2$ is given as $\theta$.
• Incident Ray ($R_{in}$): A ray of light strikes $M_1$. This ray has the property of being parallel to $M_2$.
• First Reflection: The ray reflects from $M_1$. Let the angle of incidence be $i_1$ and the angle of reflection be $r_1$. By the law of reflection, $i_1 = r_1$.
• Second Reflection: The reflected ray travels to $M_2$ and strikes it. Let the angle of incidence on $M_2$ be $i_2$ and the angle of reflection be $r_2$. By the law of reflection, $i_2 = r_2$.
• Final Ray ($R_{ref2}$): The ray reflects off $M_2$. This final ray has the property of being parallel to $M_1$.
• Objective: We need to calculate the angle of incidence $i_1$ on the first mirror $M_1$.

Geometric Setup and Triangle Relationships

To understand the relationship between the angles, let's visualize the path of the light ray and the normals to the mirrors. Consider the triangle formed by the intersection point of the mirrors (let's call it O), the point where the ray hits $M_1$ (point A), and the point where the ray hits $M_2$ (point B). This forms $\triangle OAB$.

• The angle at vertex O, $\angle AOB$, is the angle between the mirrors, so $\angle AOB = \theta$.
• At point A on $M_1$, the angle inside the triangle, $\angle OAB$, is the angle between the ray segment AB and the mirror $M_1$. Since $r_1$ is the angle of reflection (measured from the normal to $M_1$), the angle between the ray AB and the mirror $M_1$ is $90^\circ - r_1$. As $r_1 = i_1$, we have $\angle OAB = 90^\circ - i_1$.
• At point B on $M_2$, the angle inside the triangle, $\angle OBA$, is the angle between the ray segment AB and the mirror $M_2$. The angle of incidence on $M_2$ is $i_2$. The angle between the ray AB and the mirror $M_2$ is $90^\circ - i_2$. So, $\angle OBA = 90^\circ - i_2$.

The sum of the angles in $\triangle OAB$ must be $180^\circ$: $ \angle AOB + \angle OAB + \angle OBA = 180^\circ $ Substituting the angle expressions: $ \theta + (90^\circ - i_1) + (90^\circ - i_2) = 180^\circ $ Let's simplify this equation: $ \theta + 180^\circ - i_1 - i_2 = 180^\circ $ $ \theta - i_1 - i_2 = 0 $ Rearranging gives: $ i_1 + i_2 = \theta \quad (*) $ This fundamental relationship connects the two incidence angles and the mirror angle $\theta$.

Using the Parallel Ray Conditions

The problem provides two crucial conditions regarding parallel rays. Let's translate these into equations:

1. Incident Ray $R_{in}$ is parallel to $M_2$: If a ray is parallel to a mirror, it makes an angle of $0^\circ$ with that mirror. When this ray ($R_{in}$) hits $M_1$, the angle it makes with $M_1$ can be related to the angle between the mirrors, $\theta$. Specifically, the angle between the incident ray and $M_1$ is equal to $\theta$. We know the angle of incidence $i_1$ relates to the angle the ray makes with the mirror $M_1$ as $90^\circ - i_1$. Therefore, we set the angle with the mirror equal to $\theta$: $ 90^\circ - i_1 = \theta $ Solving for $i_1$: $ i_1 = 90^\circ - \theta \quad (1) $
2. Final Ray $R_{ref2}$ is parallel to $M_1$: Similarly, the final ray ($R_{ref2}$) makes an angle of $0^\circ$ with $M_1$. As this ray leaves mirror $M_2$, the angle it makes with $M_2$ must be equal to the angle between the mirrors, $\theta$. The angle of reflection $r_2$ (which is equal to $i_2$) relates to the angle the ray makes with mirror $M_2$ as $90^\circ - r_2$, or $90^\circ - i_2$. Equating this angle to $\theta$: $ 90^\circ - i_2 = \theta $ Solving for $i_2$: $ i_2 = 90^\circ - \theta \quad (2) $

Solving for the Angle of Incidence $i_1$

We have derived three equations:

• $i_1 + i_2 = \theta$ (from the geometry of $\triangle OAB$)
• $i_1 = 90^\circ - \theta$ (from the incident ray condition)
• $i_2 = 90^\circ - \theta$ (from the final ray condition)

Now, we substitute equations (1) and (2) into the geometric equation ($*$): $ (90^\circ - \theta) + (90^\circ - \theta) = \theta $ Combine terms involving $\theta$: $ 180^\circ - 2\theta = \theta $ Add $2\theta$ to both sides to solve for $\theta$: $ 180^\circ = 3\theta $ $ \theta = \frac{180^\circ}{3} $ $ \theta = 60^\circ $

This calculation shows that the conditions given in the problem are only possible when the mirrors are inclined at an angle of $60^\circ$. The question asks for the angle of incidence on the first mirror, $i_1$. We can find this using equation (1): $ i_1 = 90^\circ - \theta $ Substitute the value of $\theta$ we found: $ i_1 = 90^\circ - 60^\circ $ $ i_1 = 30^\circ $

Therefore, the angle of incidence on the first mirror ($M_1$) is $30^\circ$. This matches the second option provided.