# Calculate the image distance for an object of height 12 mm at a distance of 0.20 m from a concave lens of focal length 0.30 m, and state the nature and size of the image.

Given:

Focal length of concave lens, $f$ = $-$0.30 m

​Object distance, $u$ = $-$0.20 m                         (object distance is always taken negative, as it is placed on the left side of the lens)

Object height, $h$ = $+$12 mm = $+$0.012 m

To find: Nature and 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 in the formula, we get-

$\frac {1}{v}-\frac {1}{(-0.20)}=\frac {1}{(-0.30)}$

$\frac {1}{v}+\frac {1}{0.20}=-\frac {1}{0.30}$

$\frac {1}{v}=-\frac {1}{0.30}-\frac {1}{0.20}$

$\frac {1}{v}=-\frac {100}{30}-\frac {100}{20}$

$\frac {1}{v}=\frac {-200-300}{60}$

$\frac {1}{v}=-\frac {500}{60}$

$v=-\frac {60}{500}$

$v=-0.12m$

Thus, the image is at a distance of 0.12 m from the concave lens, and the negative sign implies that it is on the left side of it. Therefore, the image formed is virtual.

Now,

Fro magnification we know that-

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

Substituting the given values in the formula, we get-

$\frac {-0.12}{-0.20}=\frac {h'}{0.012}$

$\frac {12}{20}=\frac {h'}{0.012}$

$\frac {3}{5}=\frac {h'}{0.012}$

$h'=\frac {3\times {0.012}}{5}$

$h'=\frac {0.036}{5}$

$h'=+0.0072m=+7.2mm$

Thus, the size of the image $h'$ is 7.2 mm, and the positive sign implies that the image is erect. Also, the size of the image is smaller than the size of the object, so it is diminished.

Therefore, the nature of the image formed by the concave lens is virtual, erect, and diminished.

Updated on: 10-Oct-2022

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