Question

An object 50 cm tall is placed on the principal axis of a convex lens. Its 20 cm tall image is formed on the screen placed at a distance of 10 cm from the lens. Calculate the focal length of the lens. ​

Answer 1

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Given :

• height of the object = 50 cm

• height of the image = - 20 cm (real )

• image distance = 10 cm

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To find :

• focal length

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Things we need to know :

• object distance is always negative as distance is measured from the pole if the lens
• In this case image distance is positive
• focal length of a convex lens is always positive
• real images are always inverted

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Concept :

• We need to find the object distance first by using the magnification formula
• Once we obtain the object distance we will substitute it's value along with the image distance into the lens formula in order to obtain the focal length

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Solution :

Magnification is given by,

$\star \boxed{ \mathtt{m = \dfrac{v}{u} = \dfrac{h_i}{h_o} }}$

here ,

• m = magnification
• v = image distance
• u = object distance
• hi = height of image
• ho = height of object

Substituting the given values in the above equation ,

$\implies \mathtt{u = v \times \dfrac{h_o}{h_i} }$

$\implies \mathtt{u = 10\times \dfrac{50}{ - 20} }$

$\implies \mathtt{u = - 25 \: cm}$

The image distance is - 25 cm

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By using the lens formula ,

$\star \boxed{ \mathtt{ \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}}}$

here ,

• f = focal length
• v = image distance
• u = object distance

Substituting the given values in the above equation ,

$\implies \mathtt{\dfrac{1}{f} = \dfrac{1}{10} - \dfrac{1}{ - 25}}$

$\implies \mathtt{\dfrac{1}{f} = \dfrac{1}{10} + \dfrac{1}{ 25}}$

$\implies \mathtt{\dfrac{1}{f} = \dfrac{5 + 2}{50}}$

$\implies \mathtt{f = \dfrac{50}{7} }$

$\implies \mathtt{ \red{f = 7.14 \: cm}}$

The focal length of the lens is 7.14 cm