A point source of light S, placed at a distance 60cm in front of the centre of a plane mirror of width 50 cm, hangs vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a distance 1.2 m from it (see in the figure). The distance between the extreme points where he can see the image of the light source in the mirror is \underline\hspace2cm cm.

Geometrical Optics PYQ — JEE Main 2021 · 26 Feb · Shift 2 · Q73 with Solution

View written solutionFree

Correct answer: 150

A plane mirror forms the virtual image $S'$ of the source $S$ at the same distance behind the mirror as the source is in front.

Given:

Source distance from mirror $= 60\text{ cm}$
Man walks along a line parallel to mirror at distance $1.2\text{ m} = 120\text{ cm}$
Mirror width $= 50\text{ cm}$

So the image $S'$ is $60\text{ cm}$ behind the mirror.

Hence the distance between the man's path and the image is $120 + 60 = 180\text{ cm}.$

A man at position $P$ can see the image $S'$ if the straight line joining $P$ to $S'$ meets the plane of the mirror within its width.

Let the mirror extend from $y=-25\text{ cm}$ to $y=+25\text{ cm}$ , with its center at $y=0$ .

Let the man's position on his path be $y=y_P$ . The source/image is opposite the center, so $S'$ has coordinate $y=0$ .

The line joining $P(y_P)$ and $S'(0)$ cuts the mirror plane in the ratio $\frac{120}{180} = \frac{2}{3}$ from $P$ toward $S'$ .

Therefore the $y$ -coordinate of intersection at the mirror is

= y_P\left(1-\frac{2}{3}\right) = \frac{y_P}{3}.$$ For the image to be visible, $$|y_m| \le 25.$$ So, $$\left|\frac{y_P}{3}\right| \le 25 \quad \Rightarrow \quad |y_P| \le 75.$$ Thus the man can see the image while moving between $$y_P=-75\text{ cm} \quad \text{and} \quad y_P=+75\text{ cm}.$$ 3. **Distance between extreme points** Hence the distance between the extreme positions is $$75-(-75)=150\text{ cm}.$$ 4. **Final answer** The required distance is $$\boxed{150\text{ cm}}.$$