Find the distance $(u)$ of an object, of height $h_{o} = 5 c m$ from the optical centre of a concave lens, so that its image is formed $15 c m$ away from lens? Also, find the size of the image ( $h_{i}$ )?
Focal length of concave lens, $f = 30 c m$ .

u = 30 c m

h_{i} = - 2.5 c m

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u = - 30 c m

h_{i} = 2.5 c m

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u = 10 c m

h_{i} = 5 c m

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u = - 10 c m

h_{i} = 5 c m

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Solution

The correct option is B $u = - 30 c m$
$h_{i} = 2.5 c m$
Given:
Focal length of concave lens, $f = - 30 c m$
Height of object, $h_{o} = 5 c m$
Image distance, $v = - 15 c m$
Object distance is $u$
Image height is $h_{i}$
(Assuming new cartesian sign convention)

Using lens formula,
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{- 30} = \frac{1}{(- 15)} - \frac{1}{u}$
$u = - 30 c m$

Using magnification formula,
$m = \frac{h_{i}}{h_{o}} = \frac{v}{u}$
$\frac{h_{i}}{5} = \frac{- 15}{- 30}$
$h_{i} = 2.5 c m$

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Find the distance (u) of an object, of height ho=5 cm from the optical centre of a concave lens, so that its image is formed 15 cm away from lens? Also, find the size of the image (hi)? Focal length of concave lens, f=30 cm.

Find the distance $(u)$ of an object, of height $h_{o} = 5 c m$ from the optical centre of a concave lens, so that its image is formed $15 c m$ away from lens? Also, find the size of the image ( $h_{i}$ )?
Focal length of concave lens, $f = 30 c m$ .