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

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:

• The object shoulkd be placed at a distance of 36 cm from the mirror.

• The magnification of the image will be 1.

Given:

Focal length(f) = 18 cm.

The distance of the image(v) = 36 cm.

To find:

• The distance of the object.
• The magnificent of the image.

Explanation:

Let, the distance of the object be 'u' from the mirror.

We know that,

$\frac{1}{f} = (\frac{1}{u} + \frac{1}{v})$

⇒ $\frac{1}{18} = (\frac{1}{u} + \frac{1}{36} )$

⇒ $(\frac{1}{18} - \frac{1}{36} ) = \frac{1}{u}$

⇒ $\frac{1}{u}= (\frac{1}{18} - \frac{1}{36})$

⇒ $\frac{1}{u} = (\frac{2-1}{36} )$ [∵ The L.C.M. of the denominators is 36.]

⇒$\frac{1}{u}= \frac{1}{36}$

⇒ u = 36 cm.

∴ The distance of the object from the mirror is 36 cm.

Let, the magnificent of the image be, 'm'

We know that,

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

   $= \frac{36}{36}$

   = 1

∴ The magnificent of the image is 1.

Since, magnificent has no unit as it is a ratio.

Answer 2

$\: \red{\boxed{\bf{Solution. }}}$

$\rm{Given, \: focal \: lenth \: of \: lens,}$

$\sf \red{f = 18 \: cm }$

$\rm{Distance \: of \: image,}$

$\sf \red{v = ± 36 \: cm}$

$\sf{ [∵ ± sign \: as \: an \: image \: can \: be \: on \: either \: side} \\ \sf{of \: lens.]}$

$\rm{Distance \: of \: object,}$

$\sf \blue{u = ?}$

$\rm{Magnification \: produced,}$

$\sf \red{m = ?}$

$\rm{From \: lens \: formula,}$

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

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

$\sf \: = \frac{1}{±36} - \frac{1}{18} = \frac{1}{36} - \frac{1}{18} \sf \frac{1}{ - 36} - \frac{1}{18}$

$\sf = \frac{1 - 2}{36}\ and \frac{ - 1 - 2}{36} = \frac{ - 1}{36} \: and \frac{ - 3}{36}$

$\sf{ ∴ u = - 36 \: cm \: and \: - 12 \: cm}$

$\rm \: If \: \: \: \: \: \: \: \ \: \: \: \: \sf{u = - 36 \: cm \: and \: v = 36 \: cm}$

$\rm Then, \: \: \: \: \: \: \: \: \: \: \sf \red{m = \frac{v}{u} }$

$\sf \frac{36}{ - 36} = \bf \red{ - 1}$

$\bf \pink{Image \: is \: real \: and \: inverted.}$

$\rm \: If \: u \: = \sf \red{- 12 \: cm} \: and \: v = \sf \red{ - 36 \: cm}$

$\rm Then, \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: m = \frac {v}{u}$

$\rm = \frac{ - 36}{ - 12}$

$\sf \ = \red{3 }$

$\rm Image \:is \: virtual \: and \: erect.$