An object of 5.0 cm size is placed at a distance of 20.0 cm from a converging mirror of focal length 15.0 cm. At what distance from the mirror should a screen be placed to get the sharp image? Also, calculate the size of the image.

Given:

A concave mirror is a converging mirror.

Distance of the object from the mirror $u$ = $-$20 cm

Height of the object, $h_{1}$ = 5 cm

Focal length of the mirror, $f$ = $-$15 cm

To find: Distance of the image $(v)$ from the mirror, and the height of the image $(h_2)$.

Solution:

From the mirror formula, we know that-

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

Substituting the given values in the mirror formula we get-

$\frac{1}{(-15)}=\frac{1}{v}+\frac{1}{(-20)}$

$-\frac{1}{15}=\frac{1}{v}-\frac{1}{20}$

$\frac{1}{20}-\frac{1}{15}=\frac{1}{v}$

$\frac{1}{v}=\frac{3-4}{60}$

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

$v=-60cm$

Thus, the distance of the image, $v$ is 60 cm from the mirror, and the negative sign implies that the image forms in front of the mirror (on the left).

Now, from the magnification formula, we know that-

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

Substituting the given values in the magnification formula we get-

$\frac{{h}_{2}}{5}=-\frac{(-60)}{(-20)}$

$\frac{{h}_{2}}{5}=-3$

$h_2=5\times {(-3)}$

$h_2=-15cm$

Thus, the height of the image $h_2$ is 15 cm, and the negative sign implies that the image forms below the principal axis (downwards).

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Updated on: 10-Oct-2022

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