Question

A covex lens produces a double size real image when an object is placed at a distance of 18cm from it. The object be placed to produce a triple size real image is____

Answer 1

Hope u like my process
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The formula of lens is given as,

$= > \bf \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
It can be simplified as,

$= > \bf \: u(\frac{1}{v} - \frac{1}{u} ) = \frac{u}{f} \\ \\ = > \frac{u}{v} - \frac{u}{u} = \frac{u}{f} \\ \\ = > \frac{1}{m} = \frac{u}{f} + 1 = \frac{u + f}{f} \\ \\ = > \bf \: m = \frac{f}{f + u}$

Where

m is the magnification ;

f is focal length ;

u is the position of object.

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Using sign convention for convex lens,
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f = (+) ve ; u = (-) ve
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Thus, by problem
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U = - 18 cm ; m = 2
$= > 2 = \frac{f}{f + ( - 18)} \\ \\ = > 2f - 36 = f \\ \\ = > 2f - f = 36 \\ \\ = > \bf \: f \: = \underline{ \: 36 \: \: cm \: \: }$

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Now,

since the same lens is used

So, f will be the same.
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Thus if,

m = 3 & f = 36 ,

Then the required u can be calculated as

$= > 3 = \frac{36}{36 + ( - u)} \\ \\ = > 36 - u = \frac{36}{3} \\ \\ = > \bf \: u = 36 - 12 = \underline{ \: 24 \: \: cm \: \: }$

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Thus the required position of the object is 24 cm.

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