Two plane mirrors {M}_{1} and {M}_{2} are at | vedclass

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(B) Let the corner where the mirrors meet be the origin $(0,0)$. The position of the point source $P$ is $(a, 2a)$.The image formed by mirror ${M}_{1}

Vedclass · Accepted Answer

$4.6a$

Vedclass · Answer

(B) Let the corner where the mirrors meet be the origin $(0,0)$. The position of the point source $P$ is $(a, 2a)$.The image formed by mirror ${M}_{1}$ (at $x=0$) is ${I}_{1} = (-a, 2a)$.The image formed by mirror ${M}_{2}$ (at $y=0$) is ${I}_{2} = (a, -2a)$.The image formed by both mirrors (the intersection of the reflections) is ${I}_{3} = (-a, -2a)$.We need to find the distances between all pairs of images:$1$. Distance between ${I}_{1}$ and ${I}_{2}$: $\sqrt{(a - (-a))^2 + (-2a - 2a)^2} = \sqrt{(2a)^2 + (-4a)^2} = \sqrt{4a^2 + 16a^2} = \sqrt{20a^2} = 2\sqrt{5}a = 2 	imes 2.3a = 4.6a$.$2$. Distance between ${I}_{1}$ and ${I}_{3}$: $\sqrt{(-a - (-a))^2 + (-2a - 2a)^2} = \sqrt{0 + (-4a)^2} = 4a$.$3$. Distance between ${I}_{2}$ and ${I}_{3}$: $\sqrt{(-a - a)^2 + (-2a - (-2a))^2} = \sqrt{(-2a)^2 + 0} = 2a$.The longest distance is $4.6a$. Thus,the correct option is $B$.

Two plane mirrors ${M}_{1}$ and ${M}_{2}$ are at a right angle to each other as shown. $A$ point source $P$ is placed at distances of $a$ and $2a$ meters from ${M}_{1}$ and ${M}_{2}$ respectively. The longest distance between the images thus formed is: (Take $\sqrt{5} = 2.3$)

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