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-

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}