the image of needle placed 10 cm from a lens in front a lens is formed on a wall 20 cm on the other side of the lens. find the focal length of the lens and the size of image formed if size of object needle is 2.5 cm

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Solution

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$d i s \tan c e o f t h e o b j e c t f r o m l e n s, u = - 10 c m (- s i g n d u e t o d i r e c t i o n o p p o s i t e t o i n c i d e n t r a y) d i s \tan c e o f t h e i m a g e = 20 c m \frac{1}{v} - \frac{1}{u} = \frac{1}{f}, \frac{1}{f} = \frac{1}{20} - (- \frac{1}{10}) f = \frac{20}{3} c m, i m a g e f o r m e d i s e n l a r g e d, r e a l a n d i n v e r t e d . m a g n i f i c a t i o n, m = \frac{v}{u} = \frac{s i z e o f t h e i m a g e}{s i z e o f o b j e c t} s i z e o f i m a g e = - \frac{20}{10} \times 2.5 = - 5 c m (- s i g n i s d u e t o r e a l a n d i n v e r t e d i m a g e) R e g a r d s$

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