An object of height 5 cm is placed perpendicular to the principal axis of a concave lens of focal length 10 cm. Use lens formula to determine the position, size, and nature of the image, if the distance of the object from the lens is 20 cm.

Given:

Object height, $h=5cm$

Focal length, $f=-10cm$

Object distance, $u=-20cm$

Applying the lens formula:

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

$\therefore \frac{1}{v}=\frac{1}{f}+\frac{1}{u}$

Substituting the given value we get-

$\frac{1}{v}=\frac{1}{-10}+\frac{1}{-20}$

$\frac{1}{v}=\frac{1}{-10}-\frac{1}{20}$

$\frac{1}{v}=\frac{-2-1}{20}$

$\frac{1}{v}=\frac{-3}{20}$

$v=-\frac{20}{3}$

$v=-6.67cm$

Hence the image is formed 6.67cm in front of the lens and on the same side as the object. As $v$ is negative, we can say that the image is virtual.

Now height of image can be calculated from the mgnification formula for the lens: 

$m=\frac{{h}_{2}}{{h}_{1}}=\frac{v}{u}$

$\therefore {h}_{2}=\frac{v{h}_{1}}{u}$

Putting the given value we get-

${h}_{2}=\frac{-6.67\times 5}{-20}$

${h}_{2}=1.67cm$

Hence, the size of the image is 1.67cm

As the height of the image is positive and smaller than the height of the object, the image is erect and diminished.

So, we can conclude that the image is virtual, erect and diminished.

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Updated on: 10-Oct-2022

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