Question

A convex lens forms an erect & three times magnified image of an object placed at a distance 10 cm in front of it. Find a) position of image b) the focal length of lens​

Answer 1

Answer:

A) the object is placed between f1and optic center

. b) the focal length could be found through lens formulae as magnification is +3.

+3u=v

1/f=1/v-1/u

=1/3u-1/u

=(1-3)/3u

=-2/3*-20(u =-20)

=f=15cm

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Answer 2

Given:

• Convex lens forms erect and 3x magnified image.
• Object is placed 10 cm in front of lens.

To find:

• Image distance
• Focal length of lens

Calculation:

For lens, magnification is written as :

$\therefore \: mag. = \dfrac{ h_{i}}{ h_{o} } = \dfrac{v}{u}$

$\implies \: 3 = \dfrac{v}{u}$

$\implies \: v = 3u$

$\implies \: v = 3 \times ( - 10)$

$\implies \: v = - 30 \: cm$

So, image is formed 30 cm in front of lens.

Now, applying LENS FORMULA:

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

$\implies \dfrac{1}{f} = \dfrac{1}{( - 30)} - \dfrac{1}{( - 10)}$

$\implies \dfrac{1}{f} = - \dfrac{1}{ 30} + \dfrac{1}{ 10}$

$\implies \dfrac{1}{f} = \dfrac{ - 1 + 3}{ 30}$

$\implies \dfrac{1}{f} = \dfrac{ 2}{ 30}$

$\implies f = 15 \: cm$

So, focal length of lens is 15 cm.