An object 60cm tall is placed on the principal axis of a convex lens and its 30cm tall image is formed on the screen placed at a distance of 40cm from the object.What is the focal length of the lens?


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Question Number 51637 by Necxx last updated on 29/Dec/18

$A n o b j e c t 60 c m t a l l i s p l a c e d o n$ $t h e p r i n c i p a l a x i s o f a c o n v e x l e n s$ $a n d i t s 30 c m t a l l i m a g e i s f o r m e d$ $o n t h e s c r e e n p l a c e d a t a d i s t a n c e$ $o f 40 c m f r o m t h e o b j e c t . W h a t i s$ $t h e f o c a l l e n g t h o f t h e l e n s ?$
	Answered by aseerimad last updated on 29/Dec/18

	Commented by aseerimad last updated on 29/Dec/18
	pls verify...i am also a student of this level....
	Commented by Kunal12588 last updated on 29/Dec/18

	$m e t o o (c l a s s 11)$
	Commented by aseerimad last updated on 29/Dec/18
	is it ok?
	Answered by tanmay.chaudhury50@gmail.com last updated on 29/Dec/18

	$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ $1 + \frac{v}{u} = \frac{v}{f}$ $1 + m = \frac{v}{f}$ $u + v = d$ $u (1 + \frac{v}{u}) = d$ $1 + m = \frac{d}{u}$ ${(1 + m)}^{2} = \frac{v}{f} \times \frac{d}{u} = \frac{m d}{f}$ $f = \frac{m d}{{(1 + m)}^{2}} = \frac{\frac{1}{2} \times 40}{{(1 + \frac{1}{2})}^{2}} = \frac{20 \times 4}{9} = \frac{80}{9}$
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