Two plane mirrors are at right angles to each | vedclass

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(B) When two plane mirrors are placed at an angle of $90^{\circ}$,the number of images formed is given by $n = (360^{\circ}/	heta) - 1 = (360^{\circ}

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$1$

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(B) When two plane mirrors are placed at an angle of $90^{\circ}$,the number of images formed is given by $n = (360^{\circ}/	heta) - 1 = (360^{\circ}/90^{\circ}) - 1 = 3$ images.Let the man be at position $(x, y)$ relative to the mirrors along the $x$ and $y$ axes.The three images are formed at positions $(x, -y)$,$(-x, y)$,and $(-x, -y)$.$1$. The image at $(x, -y)$ is formed by reflection in one mirror; it undergoes lateral inversion,so the right hand appears as the left hand.$2$. The image at $(-x, y)$ is formed by reflection in the other mirror; it also undergoes lateral inversion,so the right hand appears as the left hand.$3$. The image at $(-x, -y)$ is formed by two successive reflections (one in each mirror). Each reflection causes a lateral inversion. Two lateral inversions cancel each other out,resulting in an erect image where the right hand still appears as the right hand.Therefore,in only $1$ image will he be seen using his right hand.

Two plane mirrors are at right angles to each other. $A$ man stands between them and combs his hair with his right hand. In how many of the images will he be seen using his right hand?

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