The image of the object in a plane mirror is as far behind the mirror as the object is in front of it. Therefore, the image of the boy comes near the mirror through the distance equal to that moved by the boy towards the plane mirror. Hence, the image of the boy will approach him with double his speed, i.e. with 4 m|s.

Let \(\theta \) be the angle between the mirrors \({ M }_{ 1 }\) and \({ M }_{ 2 }\)
Since, rays LM and NS are parallel to each other

\(\angle LMN=\angle SNQ=2i\)
\( \Rightarrow \quad 2i_{ 1 }+2i=180^{ \circ }\)
\(\Rightarrow i_{ 1 }+i=90^{ \circ }...............(i)\)
\(Also \ in \ \triangle MBN,\)
\( \theta +\left( 90-i \right) +\left( 90-i \right) =180^{ \circ }\)
\(\Rightarrow \theta =i_{ 1 }+i..............(ii)\)
On comparing Eqs. (i) and (ii), we get
\(​​\theta =90^{ \circ }\)

If the lower half of the concave mirror is blackened, then there is no change in the position of image, only intensity will get reduced.

A convex mirror produces a real image of a virtual object. Therefore, if a beam of light from a virtual object converges to a point behind the convex mirror, then its real image will be formed in front of the mirror.

Since, prisms can bend the light rays by 90^o and 180^o by total internal reflection so, they are used in many optical instruments.