A 3 cm high object is placed at a distance of 80 cm from a concave lens of focal length 20 cm. Find the position and size of the image.

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Solution

Here, $u = - 80 c m, f = - 20 c m$

Using the len's formula, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} - \frac{1}{- 80} = \frac{1}{- 20}$

$v = - 16 c m$

Magnification, $m = \frac{h^{^{'}}}{h} = \frac{v}{u} = \frac{- 16}{- 80} = \frac{1}{5}$

$h^{^{'}} = \frac{h}{5} = \frac{3.0}{5} = 0.6 c m$

Length of the image is 0.6 cm. Positive sign shows that the image is erect.

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A 3 cm high object is placed at a distance of 80 cm from a concave lens of focal length 20 cm. Find the position and size of the image.

Here, u=−80cm,f=−20cm Using the len's formula, 1v−1u=1f 1v−1−80=1−20 v=−16cm Magnification, m=h′h=vu=−16−80=15 h′=h5=3.05=0.6cm Length of the image is 0.6 cm. Positive sign shows that the image is erect.