Two plane mirrors M 1 and M 2 are at right angle to each other shown. A point source 'P' is placed at 'a' and '2a' meter away from M 1 and M 2 respectively. The shortest distance between the images thus formed is: (Take √5=2.3 )

Question

Two plane mirrors M 1 and M 2 are at right angle to each other shown. A point source 'P' is placed at 'a' and '2a' meter away from M 1 and M 2 respectively. The shortest distance between the images thus formed is: (Take √5=2.3 ) (A) 3 a (B) 4.6 a (C) 2.3 a (D) 2 √10 a

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/7deebefd27c0595a8fa35e21be68a520-.png" /> Shortest distance is 2 a between I 1 I 3 But answer given is for I 1 I 2 √(4 a)2+(2 a)2 a √20 4.47 a

Q. Two plane mirrors $M_{1}$ and $M_{2}$ are at right angle to each other 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 $5 = 2.3$ )

$3 a$

$4.6 a$

$2.3 a$

$210 a$

Shortest distance is $2 a$ between $I_{1} & I_{3}$
But answer given is for $I_{1} & I_{2}$
$(4 a)^{2} + (2 a)^{2}$
$a 20$
$4.47 a$

Q. 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 )

3a

4.6a

2.3a

210​a