Question

An object is placed vertically at a distance of 20 cm from a convex lens. If the height of the object is 5 cm and the focal length of the lens is 10 cm, what will be the position, size and nature of the image? How much bigger will the image be as compared to the object?​
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Answer 1

Given :-

Height of the object (h¹) = 5cm,

focal length (f) = 10cm, distance of the object ( u ) = -20cm

Image distance (v) = ?,

Height of the image ( h²) = ?,

⠀⠀⠀⠀⠀⠀⠀ Magnification (M) = ?

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

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

$\sf \longrightarrow\: \dfrac{1}{v} = \dfrac{1}{ -20} + \dfrac{1}{10}$

$\sf \longrightarrow\: \dfrac{1}{v} - \dfrac{ - 1 + 2}{20}$

$\sf \longrightarrow\: \dfrac{1}{v} = \dfrac{1}{20}$

$\sf \longrightarrow\: v \: = 20cm$

The positive sign of the image distance shows that image is formed at 20cm on the other side of the lens.

$Magnification(M) = { \bf \dfrac{h_2 }{ \ h_1 }} = \bf\dfrac{v}{u}$

$h_2 = { \bf \dfrac{v}{ \ u }} \times h_1$

$h_2 = { \bf \dfrac{20}{ - 20 }} \times 5$

$h_2 = { \bf {( - 1) \times 5}{ }}$

$h_2 = { \bf { - 5cm}{ }}$

$M = { \bf \dfrac{v}{ u }} = \dfrac{20}{ - 20} = - 1$

The negative sign of the height of image and the magnification shows the image in inverted and real. it is below the principal axis and is same size as the object.