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D e a r S t u d e n t, H e r e, w e c o n s i d e r b o t h c o n c a v e a n d c o n v e x m i r r o r s . I m a g e i s a l w a y s d i m i n i s h e d a s m a g n i f i c a t i o n, m < 1 C o n c a v e m i r r o r : m a g n i f i c a t i o n, m = - \frac{1}{3}, f = - 20 c m m = - \frac{v}{u} = - \frac{1}{3} \Rightarrow v = \frac{u}{3} . . . . . . . . (1) W e h a v e, \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \Rightarrow - \frac{1}{20} = \frac{1}{u} + \frac{3}{u} = \frac{4}{u} (U \sin g e q u a t i o n 1) \Rightarrow u = - 80 c m O b j e c t i s p l a c e d a t 80 c m i n f r o n t o f t h e m i r r o r . I m a g e d i s \tan c e, v = - \frac{80}{3} c m I m a g e i s f o r m e d a t \frac{80}{3} c m i n f r o n t o f t h e m i r r o r . (S a m e s i d e a s o b j e c t .) I m a g e i s r e a l a n d i n v e r t e d . C o n v e x m i r r o r : m = \frac{1}{3}, f = 20 c m m = - \frac{v}{u} = \frac{1}{3} \Rightarrow v = - \frac{u}{3} . . . . . . (2) W e h a v e, \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \Rightarrow \frac{1}{20} = \frac{1}{u} - \frac{3}{u} = - \frac{2}{u} (U \sin g e q u a t i o n 2) \Rightarrow u = - 40 c m O b j e c t i s p l a c e d a t 40 c m i n f r o n t o f t h e m i r r o r . I m a g e d i s \tan c e, v = - (\frac{- 40}{3}) = + 40 c m I m a g e i s f o r m e d a t \frac{40}{3} c m b e h i n d t h e m i r r o r . (o p p o s i t e s i d e a s o b j e c t .) I m a g e i s v i r t u a l a n d e r e c t . R e g a r d s

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