Question

A 5.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20

cm. The distance of the object from the lens is 30 cm. By calculation determine

(i) the position and (ii) the size of the image formed.​

Answer 1

$★\large\bold{\underline{\rm{\blue{Given :-}}}}$

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• $\small\bold{\sf{u=-30cm}}$
• $\small\bold{\sf{h_{object}=5cm}}$
• $\small\bold{\sf{f=20cm...(covex\:lens)}}$

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$★\large\bold{\underline{\rm{\blue{To\: Find :-}}}}$

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• $\small\bold{\sf{v= ?? }}$
• $\small\bold{\sf{h_{image} = ?? }}$

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$\sf {\underline{According \: to \: the \: question:-}}$

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From the lens formula we get,

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

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$\implies\sf\ \: \dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}$

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$\implies\sf \: \dfrac{1}{v} = \dfrac{1}{20} + \dfrac{-1 }{30}$

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$\implies\sf \: \dfrac{1}{v} = \dfrac{ 3 - 2 }{60}$

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$\implies\sf \: \dfrac{1}{v} = \frac{ 1 }{ 60 }$

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$\implies\sf \: v = 60 cm$

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The image is formed at a distance of 60cm behind the convex lens . Therefore image is real and inverted.

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$\implies\sf \: m = - \dfrac{v}{u} \\ \implies\sf \: m = \frac{60}{ - 30}$

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$\implies\sf \: m = -2$

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Also,

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$\implies\sf \: m = \dfrac{h_{image} }{h _{object}}$

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$\implies\sf \: - 2 = \frac{h _{i} }{5}$

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$\implies\sf \: h _{i} = -2 \times 5 = -10 cm$

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Thus, height of an image is 10 cm.

Therefore,we can say that the image is enlarged .

Answer 2

Position of Image = 60 cm behind the lens

Size of image formed = 10 cm

Explanation:

Given:-

• Object distance (u) = -30 cm

• Focal length (f) = 20 cm

• Height of Object (h) = 5 cm

To Find:-

• Position of Image formed (v)

• Size of the Image formed (h')

Formula used:-

• Lens Formula:- $\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}$

• Magnification:- $m=\dfrac{v}{u}=\dfrac{h'}{h}$

Solution:-

Using Mirror Formula,

$⟹\:\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}$

$⟹\:\dfrac{1}{v}=\dfrac{1}{20}+\dfrac{1}{-30}$

$⟹\:\dfrac{1}{v}=\dfrac{3+(-2)}{60}$

$⟹\:\dfrac{1}{v}=\dfrac{3-2)}{60}$

$⟹\:\dfrac{1}{v}=\dfrac{1}{60}$

So, v = 60 cm

Now, Using Magnification Formula

$⟹\:m=\dfrac{v}{u}=\dfrac{h'}{h}$

$⟹\:\dfrac{60}{-30}=\dfrac{h'}{5}$

$⟹\:h'=5×(-2)$

$⟹\:h'=-10cm$

Hence, The Distance of the Image formed is 60 cm and the height of the image formed is 10cm.

About Image formed:-

• Real
• Inverted
• Enlarged
• Behind the lens

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