Two plane mirrors M₁ and M₂ are at right angle to each other shown. A point source 'P' is placed at 'a' and '2a' meter away from M₁ and M₂ respectively. The shortest distance between the images thus formed is : (Take √5 = 2.3)

Solution

Physics sir · 4 steps

Put the corner at the origin, reflect P in each mirror, and use Pythagoras on the two images.

1. Set up axes. Take the corner as the origin, with M₁ along the y-axis and M₂ along the x-axis. P is a from M₁ and 2a from M₂, so P is at (a, 2a).
2. Image in each mirror. A plane mirror puts the image as far behind it as the object is in front. Reflecting in M₁ flips the x-coordinate, giving I₁ = (-a, 2a). Reflecting in M₂ flips the y-coordinate, giving I₂ = (a, -2a).
3. Separation. These are the two images formed by the two mirrors. They are 2a apart horizontally and 4a apart vertically, and the straight line joining them is the shortest distance.

   I₁I₂ = √((2a)² + (4a)²) = √20 a = 2√5 a

4. So. With √5 = 2.3, the distance is 2 × 2.3 a = 4.6a, which is option (b).

The idea to keep. For mirrors at right angles, use coordinates: each mirror flips one coordinate of the object. The distance from an object to its image is twice its distance from the mirror, never just once.

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