To get three images of a single object, one should have two plane mirrors at an angle of:

30^{o}

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50^{o}

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60^{o}

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90^{o}

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Solution

The correct option is D $90^{o}$
Option (D) ie. $90^{o}$ , we get $n = \frac{360}{0} - 1$
Here $\frac{360^{o}}{90^{o}} - 1 = 4 - 1 = 3$
$\frac{360^{o}}{90^{o}} - 1 = 4 - 1 = 3$

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60^{o}

Q TO GET NINE IMAGES OF A SINGLE OBJECT,WHEN PLACED ASYYMETRICAL BETWEEN TWO PLANE MIRRORS.ANGLE BETWWEN 2 PLANE MIRRORS SHOULD BE.

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