at what distance should an object be placed from a convex lens of foca

1.	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» 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.

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»

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.

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.