Question

An object is placed at the distance of 10 cm from a concave lens lens of focal length 20cm find the position of the image and its nature

Answer 1

The position of image is 6.67 cm from the pole of concave lens and image is erect, virtual and diminished.

An object is placed at the distance of 10cm from a concave lens of focal length 20cm.

We have to find the position of the image and its nature.

Given,

• Object distance, u = -10cm
• focal length of concave lens, f = -20cm

Using lens formula,

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

$\implies\frac{1}{-20}=\frac{1}{v}-\frac{1}{-10}\\\\\implies\frac{1}{v}=\frac{1}{-20}-\frac{1}{10}\\\\\implies\frac{1}{v}=\frac{-3}{20}\\\\\implies v=-\frac{20}{3}\approx-6.67cm$

Therefore the position of image is 6.67cm from the pole of concave lens.

Magnification is the ratio of image position to object position.

$i.e., m=\frac{v}{u}$

Here, u = -10 cm , v = -20/3 cm

$\implies m=\frac{-\frac{20}{3}}{-10}=\frac{2}{3}$

Here we see that the magnification of the object is less than 1. It indicates that the size of image is smaller than that of object. and also magnification is positive, means image is erect.

Therefore the nature of image is erect, virtual and diminished.

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