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Question Number 176966 by mr W last updated on 28/Sep/22

Commented by mr W last updated on 29/Sep/22

$i s i t p o s s i b l e t h a t a l i g h t r a y s t a r t i n g$ $f r o m p o i n t O r e t u r n s b a c k t o t h e s a m e$ $p o i n t a f t e r b e i n g r e f l e c t e d a t t h e$ $p o i n t s P, Q, R o n t h e t h r e e m i r r o r$ $w a l l s ?$ $i f y e s, f i n d o u t t h e p o s i t i o n s o f t h e$ $r e f l e c t i o n p o i n t s .$ $t h e e d g e l e n g t h o f t h e c u b e i s 1 .$
	Answered by mr W last updated on 29/Sep/22

	Commented by mr W last updated on 29/Sep/22

	$P_{1} = p l a n e c o n t a i n i n g i n c i d e n t r a y$ $a n d r e f l e c t e d r a y$ $P_{2} = p l a n e c o n t a i n i n g t h e m i r r o r$ $w e h a v e :$ $P_{1} ⊥ P_{2}$ $i n c i d e n c e a n g l e = r e f l e c t i o n a n g l e = θ$
	Commented by mr W last updated on 29/Sep/22

	Commented by mr W last updated on 29/Sep/22

	$d u e t o s y m m e t r y t h e p o i n t Q m u s t l i e$ $o n t h e d i a g o n a l . i t s p o s i t i o n c a n b e$ $d e s c r i b e d t h r o u g h t h e v a l u e a .$ $B C = a$ $A D = 1 - a - a = 1 - 2 a$ $\frac{B C}{A D} = \frac{P C}{A P} = \frac{C Q}{A O}$ $\frac{a}{1 - 2 a} = \frac{a}{1}$ $\Rightarrow 1 - 2 a = 1$ $\Rightarrow a = 0$ $t h a t m e a n s i t i s n o t p o s s i b l e t h a t$ $t h e l i g h t r a y r e t u r n s t o i t s s t a r t i n g$ $p o i n t a f t e r b e i n g r e f l e c t e d b y t h r e e$ $m i r r o r s a s s h o w n .$
	Commented by Tawa11 last updated on 02/Oct/22

	$Great sir$
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