Question

an object is placed before a concave lens of focal length 12 CM the size of the image formed by the lens is half the size of the object calculate the distance of the object from the lens​

Answer 1

Answer:

- 12

Explanation:

Given An object is placed before a concave lens of focal length 12 CM the size of the image formed by the lens is half the size of the object calculate the distance of the object from the lens

We know that

1/f = 1/v – 1/u

1/v = 1/f + 1/u

1/v = f + u / fu

So v = fu / f + u

Image formed is half the size of the object.

1/2 = fu / u + f / u

Now f = - 12

1/2 = - 12 u / (u – 12) / u

So u – 12 = - 24

So u = - 12

Answer 2

Answer:

here is question from H .C . VERMA Foundation book class 10......

Explanation:

$v = \frac{uf}{u + f}$

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

m is magnification of lens

m is given 1/2

ratio of height of image of object and height of the object

therefore

$\frac{v}{u} = \frac{1}{2}$

put v = uf/u+f

therefore

$\frac{ \frac{uf}{u + f} }{u} = \frac{1}{2}$

$\frac{uf}{u + f} \times \frac{1}{u} = \frac{1}{2}$

therefore,

$\frac{f}{u + f} = \frac{1}{2}$

f = -12 cm

therefore ,

$\frac{ - 12}{u - 12} = \frac{1}{2}$

therefore

$u \: = - 12$

thank you and mark as BRAINIEST answer please