An object of height $5 c m$ is held $20 c m$ away from converging lens of focal length $10 c m$ ,Find the position, nature and size of the image formed.

v = 10 c m

, nature = real image,

h_{2} = 5 c m

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v = 10 c m

, nature = virtual image,

h_{2} = 5 c m

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v = 20 c m

, nature = real image,

h_{2} = 5 c m

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v = 20 c m

, nature = virtual image,

h_{2}

= 5 c m

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Solution

The correct option is B $v = 20 c m$ , nature = real image, $h_{2} = 5 c m$
Given, u $=$ -20cm
f $=$ 10cm
v $=$ ?
We know
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{10} = \frac{1}{v} - \frac{(- 1)}{20}$
$\frac{1}{v} = \frac{2 - 1}{20}$
$v = 20 c m$
We know
magnification (M) $= \frac{h_{i}}{h_{o}} = \frac{- v}{u}$
or, $\frac{20}{- 20} = - 1 = \frac{h_{i}}{h_{o}}$ since, $h_{0} = 5 c m$
so, $h_{i} = - 5 c m$
nature $=$ real image

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An object of height 5cm is held 20cm away from converging lens of focal length 10cm,Find the position, nature and size of the image formed.

The correct option is B v=20cm, nature = real image, h2=5cmGiven, u=-20cmf=10cmv=?We know1f=1v−1u110=1v−(−1)201v=2−120v=20cmWe knowmagnification (M)=hiho=−vuor,20−20=−1=hiho since,h0=5cmso,hi=−5cmnature = real image

An object of height $5 c m$ is held $20 c m$ away from converging lens of focal length $10 c m$ ,Find the position, nature and size of the image formed.