Question

A plane mirror is inclined at an angle of ${45}^{\circ }$ with the horizontal and having one of its end at the origin(O) as shown in figure. An object is kept at $x=-2\text{cm}$. The position coordinates of the image of $P$ will be

• A. $(+2,0)\text{cm}$
• B. $(0,+2)\text{cm}$
• C. $(0,0)\text{cm}$
• D. $(0,-2)\text{cm}$

Answer 1

The correct option is D $(0,-2)\text{cm}$

The plane mirror obeys
[Object distance = image distance]
This holds along the normal to plane mirror.


In $\Delta PON$
$\angle PON={45}^{\circ }$
$\Rightarrow \sin {45}^{\circ }=\frac{PN}{OP}$
$\Rightarrow PN=OP\times \frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\text{cm}$
Hence, $PN=N{P}^{\prime }=\sqrt{2}\text{cm}$
Again in $\Delta ON{P}^{\prime }$
$\angle NO{P}^{\prime }={45}^{\circ },\angle N{P}^{\prime }O={90}^{\circ }-{45}^{\circ }={45}^{\circ }$
$\Rightarrow ON=N{P}^{\prime }$ ($\Delta ON{P}^{\prime }$ is isosceles triangle)
Therefore,
$O{P}^{\prime }=\sqrt{O{N}^2+N{P}^{\prime 2}}$
$\Rightarrow O{P}^{\prime }=\sqrt{(\sqrt{2}{)}^2+(\sqrt{2}{)}^2}=\sqrt{4}=2\text{cm}$
Coordinates of image $P^{\prime }$ is,
$x=0,y=-2\text{cm}$
$\Rightarrow (0,-2)\text{cm}$
Hence, option $\left(d\right)$ is the correct answer.