Question

an object is placed at a distance of 60 cm from a concave lens of focal length 30 cm . find image distance using mirror formula.​

Answer 1

So I hope your board paper went well!

Also the question's supposed to be answered using lens formula not mirror.

Answer:

Given:

f = -30 (focal length is negative for concave lens)

u = -60 (object is always placed on left side of the lens/mirror)

To find:

v = ?

Lens formula:

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

Substituting using what's given:

$\frac{-1}{30} = \frac{1}{v} - \frac {-1}{60}$

$\frac{-1}{30} = \frac{1}{v} + \frac {1}{60}$

$\frac{1}{v} = \frac{-1}{30} - \frac {1}{60}$

$\frac{1}{v} = \frac{-2}{60} - \frac {1}{60}$

$\frac{1}{v} = \frac{-2-1}{60}$

$\frac{1}{v} = \frac{-3}{60}$

∴ v = $\frac{-60}{3}$

v = -20 cm

Image is virtual, erect, same side as object (20 cm away) and 1/3 of the object ($m = \frac{v}{u} = \frac{-20}{-60} = 1/3 = \frac{h'}{h}$  ∴   $h' = \frac {h}{3}$)

Answer 2

Given:

Focal length,

f = -30

The object is placed at distance,

u = -60

To Find:

Image distance, v = ?

Solution:

On applying the lens formula, we get

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

On substituting the values, we get

⇒ $-\frac{1}{30} =\frac{1}{v} +\frac{1}{60}$

⇒  $\frac{1}{v} =-\frac{1}{30} -\frac{1}{60}$

⇒  $\frac{1}{v} =\frac{-2-1}{60}$

⇒  $\frac{1}{v} =\frac{-3}{60}$

On applying cross-multiplication, we get

⇒  $60=-3v$

⇒    $v=-\frac{60}{3}$

⇒    $v=-20 \ cm$

So that the image distance will be "-20 cm".