Two vertical parallel mirrors $A$ and $B$ are separated by $10\,cm$. $A$ point object $O$ is placed at a distance of $2\,cm$ from mirror $A$. The distance of the second nearest image behind mirror $A$ from the mirror $A$ is $.........\,cm$.

$A$ beam of light from a source $L$ is incident normally on a plane mirror fixed at a certain distance $x$ from the source. The beam is reflected back as a spot on a scale placed just above the source $L$. When the mirror is rotated through a small angle $\theta$,the spot of the light is found to move through a distance $y$ on the scale. The angle $\theta$ is given by

$A$ point source of light,$S$,is placed at a distance $L$ in front of the centre of a plane mirror of width $d$ which is hanging vertically on a wall. $A$ man walks in front of the mirror along a line parallel to the mirror,at a distance $2L$ as shown below. The distance over which the man can see the image of the light source in the mirror is:

$A$ boy of height $1.5\, m$ with his eye level at $1.4\, m$ stands before a plane mirror of length $0.75\, m$ fixed on the wall. The height of the lower edge of the mirror above the floor is $0.8\, m$. Then:

Two plane mirrors ${M}_{1}$ and ${M}_{2}$ are at a right angle to each other as shown. $A$ point source $P$ is placed at distances of $a$ and $2a$ meters from ${M}_{1}$ and ${M}_{2}$ respectively. The longest distance between the images thus formed is: (Take $\sqrt{5} = 2.3$)

At what angle (in degrees) should two plane mirrors be inclined to obtain $3$ images?