Question

A point source of light, S is placed at a distance L in front of the centre of 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. $3d$
B. $\dfrac{d}{2}$
C. $d$
D. $2d$

Answer 1

The law of reflection states that when the light reflects from the plane mirror, the angle of reflection equals the angle of incidence. Draw the ray diagram of the image formation and use the property of similar triangles. Calculate the distance from the centre of the mirror where the image of the point source is still visible.

Complete Step by Step Answer:
We know that by the law of reflection, when the light reflects from the plane mirror, the angle of reflection equals the angle of incidence. Therefore, we can draw the ray diagram of the image formation of the point source as follows,

In the above ray diagram, when the light from point source meets at the end of the mirror M, it gets reflected with equal angle as angle of incidence and meets at point A at a distance y from the centre C. Therefore, the person can see the point source in the plane mirror if he is situated between the point A and E. From the above figure, we can see the triangles $\Delta POM$ and $\Delta MBA$ are similar triangles. Therefore, by the property of similar triangles, we can write,
$\dfrac{{AB}}{{MB}} = \dfrac{{MO}}{{PO}}$
Substituting $AB = y$, $MB = 2L$, $MO = \dfrac{d}{2}$ and $PO = L$ in the above equation, we get,
$\dfrac{y}{{2L}} = \dfrac{{d/2}}{L}$
$\Rightarrow y = d$
By the symmetry of the above figure, we can see the width of the image is,
$AE = y + \dfrac{d}{2} + \dfrac{d}{2} + y$
Substituting $y = d$ in the above equation, we get,
$AE = d + \dfrac{d}{2} + \dfrac{d}{2} + d$
$\therefore AE = 3d$

So, the correct answer is option A.

Note: While solving these types of questions, one thing that you should do is the use of the law of reflection. In the above figure, AE is not the width of the image of the point source. It is the distance over which the person can see the point source in the plane mirror irrespective of its position between A and E. The image of the point source forms at point P as shown in the above figure.