An object of height 2.5 cm is placed at a distance of 15 cm from the optical centre `O’ of a convex lens of focal length 10 cm. Draw a ray diagram to find the position and size of the image formed. Mark optical `O’, principal focus F and height of the image on the diagram.

Ray Diagram:

Explanation

 Given,

Object height = $h$ = 2.5 cm

Object distance = $u$ = $-$ 15 cm 

Focal length = $f$ = $+$ 10 cm

Image distance = $v$ = ?

Image height = $h'$ = ?

Solution:

Using the lens formula, we have-

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

It can be rearranged as-

$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$

Substituting the given values we get-

$\frac{1}{v}=\frac{1}{10}+\frac{1}{(-15)}$

$\frac{1}{v}=\frac{1}{10}-\frac{1}{15}$

$\frac{1}{v}=\frac{15-10}{150}$

$\frac{1}{v}=\frac{5}{150}$

$\frac{1}{v}=\frac{1}{30}$

$v=30cm$

Hence, the image distance is 30 cm.

Now, to find the height of the image we have-

$\frac{h\text{'}}{h}=\frac{v}{u}$

Substituting the given values we get-

$\frac{h\text{'}}{2.5}=\frac{30}{-15}$

$h\text{'}=\frac{30\times 2.5}{-15}$

$h\text{'}=-5cm$

Hence, the image height is 5 cm, and is enlarged (negative sign $-$ shows that the image is inverted).

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Updated on: 10-Oct-2022

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