Question

The focal length of a convex lens is 50cm. Find the position of the object to get 2 times real image of the object.

Answer 1

SOLUTION:-

When the image is real and erect from a convex lens, the magnification will be negative.

i.e.

$m = \frac{v}{u} = - 2$

Therefore,

Distance of image, v= -2u

Focal length= 50

So,

Distance of object u:

From lens formula using;

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\ \\ = > \frac{1}{50} = \frac{1}{ - 2u } - \frac{1}{u} \\ \\ = > \frac{1}{50} = \frac{ - 1 - 2}{2u} \\ \\ = > \frac{1}{50} = \frac{ - 3}{2u} \\ \\ = > 2u = - 3 \times 50 \\ \\ = > u = - \frac{3 \times 50}{2} \\ \\ = > u = - 3 \times 25 \\ \\ = > u = - 75cm$

So,

the distance of the object will be 75cm.

Hope it helps ☺️

Answer 2

$\it\huge\mathfrak\red{Answer}$]

Distance of image, v= -2u

Focal length= 50

So,

Distance of object u:

From lens formula using;

\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\ \\ = > \frac{1}{50} = \frac{1}{ - 2u } - \frac{1}{u} \\ \\ = > \frac{1}{50} = \frac{ - 1 - 2}{2u} \\ \\ = > \frac{1}{50} = \frac{ - 3}{2u} \\ \\ = > 2u = - 3 \times 50 \\ \\ = > u = - \frac{3 \times 50}{2} \\ \\ = > u = - 3 \times 25 \\ \\ = > u = - 75

When the image is real and erect from a convex lens, the magnification will be negative.