Question 11 of 295

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 $\sqrt{5}=2.3$)

A$2.3a$B$2\sqrt{10}a$C$4.6a$D$3a$

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Answer: (C) $4.6a$

We know that the formula of the distance between two points is, $d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$, now put the values from the above the diagram in the above formula, then the distance between ${I}_{1}$ and ${I}_{2},d=\sqrt{(4a{)}^{2}+(2a{)}^{2}}\approx 4.6a$