An object is placed at a distance of 100 cm from a converging lens of focal length 40 cm. (i) What is the nature of image ? (ii) What is the position of image ?

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Solution

Here, $u = - 100 c m$

$f = 40 c m$

Using the lens formula, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\Rightarrow \frac{1}{v} - \frac{1}{- 100} = \frac{1}{40}$

$\Rightarrow v = \frac{200}{3} = 66.67 c m$

Thus, the image is formed at a distance of 66.7 cm from the lens on the other side.

Now, magnification, $m = \frac{v}{u} = \frac{\frac{200}{3}}{- 100} = \frac{- 2}{3}$

So, the image is real, inverted and diminished.

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An object is placed at a distance of 100 cm from a converging lens of focal length 40 cm. (i) What is the nature of image ? (ii) What is the position of image ?

Here, u=−100 cm f=40 cm Using the lens formula, 1v−1u=1f ⇒1v−1−100=140 ⇒v=2003=66.67 cm Thus, the image is formed at a distance of 66.7 cm from the lens on the other side. Now, magnification, m=vu=2003−100=−23 So, the image is real, inverted and diminished.