Question

An object is initially at a distance of $200\text{cm}$ from a plane mirror. If the mirror approaches the object at a speed of $15\text{cm/s}$, then after $8\text{sec}$ the distance between object and its image will be

• A. $120\text{cm}$
• B. $160\text{cm}$
• C. $170\text{cm}$
• D. $30\text{cm}$

Answer 1

The correct option is B $160\text{cm}$


Initial distance between object and image,
$OI=2\times 200=400\text{cm}$
Relative velocity along perpendicular to the plane
$v_{o,m}=-v_{i,m}$
Taking leftwards as +ve direction,
$\Rightarrow v_o-v_m=-(v_i-v_m)$
$\Rightarrow 0-15=-v_i+15$
$\therefore v_i=+30\text{cm/s}$
In $\Delta t=8\text{s}$, the distance travelled/covered by image will be,
$S=v_i\times \Delta t=30\times 8=240\text{cm}$
Thus, the distance between object and image will be,
$O{I}^{\prime }=OI-S$
$\Rightarrow O{I}^{\prime }=400-240=160\text{cm}$, (After 8 sec).

Alternative:
Initial distance between object and mirror, $OM=200\text{cm}$
Distance covered by mirror in $8\text{s}$, $=15\times 8=120\text{cm}$
Distance between object and mirror, $O{M}^{\prime }=200-120=80\text{c}m$
Therefore distance between object and its image, $O{I}^{\prime }=2\times 80=160\text{cm}$