Question

at what distance should an object be placed from a convex lens of focal length 18 cm to obtain an image at 36 cm from it what will be the magnification produced in this case​

Answer 1

$\huge\pink{\boxed{\sf AnSwer}}$

Given:-

The focal length of the lens (f) is 18 cm

Distance of the image (v) = 36 cm

To be found

Distance of the object (u) = ?

Magnification produced (m) = ?

Now,

From the lens formula

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

$\implies \boxed{\tt \frac{1}{u}=\frac{1}{v} -\frac{1}{f} }$

So,

$\implies \tt \frac{1}{u}=\frac{1}{36} -\frac{1}{18}$

$\implies \tt \frac{1}{u}=\frac{1-2}{36}$

$\implies \tt \frac{1}{u}=\frac{-1}{36}$

So, u = -36 cm

∴ So, the distance of the object (u) = -36 cm

Now,

If u = -36

Then, we will find the magnification,

$\sf m = \frac{-v}{u} =\frac{-36}{-36} = \boxed{1}$

m = 1 so, the image is virtual and erect.

Answer 2

Answer:

By using lens formula

v

1

−

u

1

=

f

1

where v=24cm ,f=18cm

24

1

−

u

1

=

18

1

from this we get u=−72cm

magnification of lens is m=

u

v

m=

−72

24

=−

3

1