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

Since ,

Focal length=18 cm Distance of image from lens=36cm

1/f = 1/v - 1/u

--> -1/u = 1/f - 1/v

-1/u = 1/18 - 1/36

-1/u = 3/36 - 1/36

-1/u = -2/36

--->. u = 18

Magnification will be 1

Answer 2

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.

• More Information:

When the sign of the magnification is (- ve) then the image is real and inverted.

And

When the sign of the magnification is (+ ve) then the image is virtual and erect.