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-im

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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