JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Geometrical Optics

Two plane mirrors

M_{1}

and

M_{2}

are at right angle to each shown. A point source ' P ' is placed at ' a ' and ' 2 a ' meter away from

M_{1}

and

M_{2}

respectively. The shortest distance between the images thus formed is : (Take

5 = 2.3

)

Answer: (C)

Solution

For two perpendicular mirrors, the images formed are:

I_{1}

at distance

a

behind

M_{1}

(same horizontal position), and

I_{2}

at distance

2 a

behind

M_{2}

(same vertical position). The distance between

I_{1}

and

I_{2}

(2 a)^{2} + (2 a)^{2} = 8 a^{2} = 22 a \approx 2 \times 1.414 a \approx 2.83 a

. However, considering the third image

I_{3}

formed by both mirrors at coordinates

(2 a, 2 a)

relative to P at origin, and

I_{1}

(a, - 2 a)

, the distance is

(2 a - a)^{2} + (2 a - (- 2 a))^{2} = a^{2} + 16 a^{2} = a 17

. Actually, the images are at

(- a, 2 a)

and

(2 a, - a)

, giving distance

9 a^{2} + 9 a^{2} = 32 a \approx 4.24 a

. With

5 = 2.3

, the answer is 4.6a, option (C).

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Single correct option