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

Given:

Object height, $h$ = $+$5 cm

Focal length, $f$ = $-$10 cm

 Object distance, $u$ = $-$20 cm

To find: The position and nature of the image, $v$, size of the image, $h'$.

Solution:

From the lens formula, we know that-

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

Substituting the given values, we get-

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

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

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

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

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

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

$v=-6.67cm$

Thus, the position of the image, $v$ is 6.67cm away from the lens, and the minus sign $(-)$ implies that it forms in front of the lens (on the left side). So, the image is virtual.

Now, 

From the magnification formula of the lens, we know that-

$m=\frac {h'}{h}=\frac {v}{u}$

Putting the values in the formula we get-

$\frac {h'}{5}=\frac {-6.67}{-20}$

$\frac {h'}{5}=\frac {667}{20\times 100}$

$h'=\frac {667\times 5}{20\times 100}$

$h'=1.67cm$

Thus, the size of the image, $h'$ is 1.67cm.

As the size of the image is positive and less than the size of the object, the image is erect and diminished.

Thus, we can conclude that the nature of the image is virtual and erect. The size of the image is diminished.

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Updated on: 10-Oct-2022

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