Prove that if a plane mirror is rotated through an angle a, the reflected beam is rotated through an angle 2$\mathbf{\alpha }$. Show that this result is reasonable for $\mathbf{\alpha }\mathbf{=}{\mathbf{45}}^{\mathbf{\circ }}$.

Angle of rotation of mirror is$\alpha$

Angle$\alpha ={45}^{\circ }$

Here, use the concept of reflection. As the angle of incidence is the equation of the angle of reflection, the reflected ray makes an angle twice of incident angle with the incident ray.

Consider the ray coming from the source and the incident on the mirror. The angle of incidence is $\theta$ So, the angle of reflection is also $\theta$. The reflected ray makes an angle of 2$\theta$with the incident ray.

If one rotate the mirror by an angle $\alpha$the angle of incidence will increase to $\theta +\alpha$. The reflected ray is also now increases to $\theta +\alpha$. The reflected ray will make an angle with the incident ray in this case is $2(\theta +\alpha )$. This shows that the reflected ray has been rotated by an angle

If one rotates the mirror such that the angle of incidence decreased by an angle $\alpha$, then reflected ray will makes an angle of $2(\theta -\alpha )$with the incident ray. In this case also, the reflected ray is rotated by an angle of 2$\alpha$.

Hence, one can say that if a plane mirror is rotated through an angle $\alpha$the reflected beam is rotated through an angle 2$\alpha$.

Now, consider the angle of rotation of the plane mirror is

$\alpha ={45}^{\circ }$

So, the angle of rotation of the reflected ray will become

$$\begin{gathered} 2\alpha =2\times {45}^{\circ } \\ ={90}^{\circ } \end{gathered}$$

This is an acceptable angle for rotation of the reflected beam. Hence, it is proved that the result of part a is reasonable for $\alpha ={45}^{\circ }$.