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.

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