A plane mirror of length $8cm$ is moving with speed $3ms$^-1 towards a wall in the situation as shown in the figure. Size of the spot formed on the wall is $\frac{32}{x}cm$. Find the value of $x$.

Clearly, the distance of the light source from the wall is $4cm$. 
We also know that if two angles of two different triangles are equal, both these angles will subtend equal lengths on the opposite sides of these triangles.
Now, let us apply the law of reflection and the property of equal angles in triangles to each incident ray and its corresponding reflected ray. We consider three points of reflection-the first one being the extreme left point of the plane mirror, the second one being the midpoint of the plane mirror and the third one being the extreme right point of the plane mirror. Clearly, these points are $P,X$ and $M$, respectively, as can be seen from the figure given above.
Let us take the case of reflection from point $P$. The incident ray coming from the light source gets reflected and forms a spot on the wall at a distance of $4cm$ from the extended perpendicular tangent, drawn from the point of reflection. This is because the angle of incidence is equal to the angle of reflection. Also, the opposite sides subtended by these angles in triangles $APQ$ and $RPQ$ have equal lengths. Since $AQ=4cm$, it turns out that $QR$ is also equal to $4cm$.

Now, let us take the reflection from point $X$. The incident ray coming from the light source gets reflected and forms a spot on the wall at a distance of $8cm$ from the extended perpendicular tangent, drawn from the point of reflection. This is because the angle of incidence is equal to the angle of reflection. Also, the opposite sides subtended by these angles in triangles $AXY$ and $ZXY$ have equal lengths. Since $AY=8cm$, it turns out that $YZ$ is also equal to 8cm

Similarly, let us take the reflection from point $M$. The incident ray coming from the light source gets reflected and forms a spot on the wall at a distance of $12cm$ from the extended perpendicular tangent, drawn from the point of reflection. This is because the angle of incidence is equal to the angle of reflection. Also, the opposite sides subtended by these angles in triangles $AMN$ and $OMN$ have equal lengths. Since $AN=12cm$, it turns out that $NO$ is also equal to $12cm$.

Now, if we combine all these diagrams and observe carefully, we can see that the spot of the light source has a size equal to the length between $Q$ and $O$. Length between $Q$ and $O$ is given by
QO=24cm−8cm=16cm

We are given that the size of the spot is equal to ${\displaystyle \frac{32}{x}}cm$.
Equating both the above expressions together, we have
${\displaystyle \frac{32cm}{x}}=16cm\Rightarrow x=2cm$
Therefore, the value of $x$ is 2
Hence, option B is correct.