an object is placed 5cm away from a plane mirror. the mirror is rotated tbrough angle 20^0 about the point of incidence. calculate the shortest distance between the old and the new position of the image


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Question Number 133115 by aurpeyz last updated on 19/Feb/21

$a n o b j e c t i s p l a c e d 5 c m a w a y f r o m a$ $p l a n e m i r r o r . t h e m i r r o r i s r o t a t e d$ $t b r o u g h a n g l e 20^{0} a b o u t t h e p o i n t o f$ $i n c i d e n c e . c a l c u l a t e t h e s h o r t e s t$ $d i s t a n c e b e t w e e n t h e o l d a n d t h e n e w$ $p o s i t i o n o f t h e i m a g e$
	Answered by mr W last updated on 19/Feb/21

	Commented by mr W last updated on 19/Feb/21

	$A = o b j e c t$ $b l u e = m i r r o r b e f o r e$ $r e d = m i r r o r a f t e r r o t a t i o n$ $B = i m a g e b e f o r e$ $C = i m a g e a f t e r r o t a t i o n o f m i r r o r$ $a = d i s t a n c e o f o b j e c t t o m i r r o r b e f o r e$ $b = d i s t a n c e o f o b j e c t t o m i r r o r a f t e r$ $c = d i s t a n c e n e w i m a g e t o o l d i m a g e$ $c = 2 a \sin θ = 2 \times 5 \times \sin 20 ° \approx 3.42 c m$
	Commented by aurpeyz last updated on 19/Feb/21

	$i a m v e r y h a p p y . t h a n k y o u S i r$
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