An object placed 4 cm in front of a converging lens produces a real image 12 cm from the lens.(a) What is the magnification of the image?(b) What is the focal length of the lens?(c) Draw a ray diagram to show the formation of image. Mark clearly F and 2F in the diagram.

Given:

Converging lens is a convex lens.

Distance of the object from the lens, $u$ = $-$4

Distance of the image from the lens, $v$ = 12


(a) To find: Magnification of the image $m$.

Solution:

From the magnification formula we know that-

$m=\frac {v}{u}$

Substituting the given values in the formula we get-

 $m=\frac {12}{-4}$

 $m=-3$

Thus, magnification produced by the convex lens is 3.

(b) To find: Focal length of the lens $f$.

Solution:

According to the lens formula, we know that:

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

Substituting the given values in the formula we get-

$\frac {1}{12}-\frac {1}{(-4)}=\frac {1}{f}$

$\frac {1}{12}+\frac {1}{4}=\frac {1}{f}$

$\frac {1}{f}=\frac {1+3}{12}$

$\frac {1}{f}=\frac {4}{12}$

$\frac {1}{f}=\frac {1}{3}$

$f=+3cm$

Thus, the focal length $f$ of the convex lens is 3 cm.


(c) Ray diagram showing the formation of image-


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Updated on: 10-Oct-2022

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