Question

3. The focal length of a convex lens is 25 cm. At what
distance from the optical centre of the lens an object
be placed to obtain a virtual image of twice the
size?
Ans. 12,5 cm in front of the lens​

Answer 1

Answer:

u = 12.5 cm

Explanation:

convex lens of f =25 cm,

h' =2 times height of object,

u =?

we know that m =$\frac{h'}{h}$  = $\frac{v}{u}$

then,  $\frac{2h'}{h}$ = $\frac{v}{u}$,

v =2u

from lens formula,

$\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$

$\frac{1}{2u}$ - $\frac{1}{u}$ = $\frac{1}{25}$,

$\frac{1-2}{2u}$ = $\frac{1}{25}$,

$\frac{-1}{2u}$ = $\frac{1}{25}$,

2u = -25,

u = -12.5 cm

therefore  v = 2× ( -12.5)

                 v = -25 cm

                            I hope this is helpful....