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.

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Solution

Here, $u = - 4 c m$

$v = 12 c m$

(a) So, $m = \frac{v}{u} = \frac{12}{- 4} = - 3$

(b) Using lens formula, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

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

$\Rightarrow f = 3 c m$

(c) Ray diagram:

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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.

Here, u=−4 cm v=12 cm (a) So, m=vu=12−4=−3 (b) Using lens formula, 1v−1u=1f ⇒112−1−4=1f ⇒f=3 cm (c) Ray diagram:

Here, $u = - 4 c m$

$v = 12 c m$

(a) So, $m = \frac{v}{u} = \frac{12}{- 4} = - 3$

(b) Using lens formula, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

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

$\Rightarrow f = 3 c m$

(c) Ray diagram: