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 __________ cm.

Solution

Physics sir · 4 steps

Work with the image of S, not S itself: every reflected ray seems to come from the image behind the mirror.

1. Locate the image. A plane mirror forms the image S' as far behind the mirror as S is in front, so S' is 60 cm behind the mirror, opposite its centre. The man's line is 120 cm in front of the mirror, so it is 60 + 120 = 180 cm from S'.
2. Extreme rays. Every reflected ray travels as if it came straight from S' and passed through the mirror. The extreme rays pass through the two edges of the mirror, each 25 cm from the centre line. The man sees the image only between the points where these two rays meet his line.

   x/(25 cm) = (180 cm)/(60 cm) ⇒ x = 75 cm

3. Similar triangles. Here x is the distance on the man's line from the centre line to one extreme point. By symmetry the other extreme point is also 75 cm away, on the other side.
4. So. The distance between the extreme points is 2 × 75 = 150 cm.

The common slip: 100. Measuring the man's distance from the mirror (120 cm) instead of from the image (180 cm) gives 2 × 25 × 120/60 = 100 cm, not 150.

The idea to keep. For any field-of-view question with a plane mirror, draw lines from the image through the mirror's edges. Measure every distance from the image, not from the mirror or the object.

Answer from the NTA final answer key. Solution by PrepWiser.