A bright object 50 mm high stands on the axis of a concave mirror of focal length 100 mm and at a distance of 300 mm from the concave mirror. How big will the image be?

Given:

Distance of the object from the mirror $u$ = $-$300 cm

Height of the object, $h_{1}$ = 50 mm

Focal length of the mirror, $f$ = $-$100 mm

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-

$\frac{1}{(-100)}=\frac{1}{v}+\frac{1}{(-300)}$

$-\frac{1}{100}=\frac{1}{v}-\frac{1}{300}$

$\frac{1}{300}-\frac{1}{100}=\frac{1}{v}$

$\frac{1}{v}=\frac{1-3}{300}$

$\frac{1}{v}=\frac{-2}{300}$

$\frac{1}{v}=\frac{-1}{150}$

$v=-150mm$

Thus, the distance of the image, $v$ is 150 mm 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-

$\frac{{h}_{2}}{50}=-\frac{(-150)}{(-300)}$

$\frac{{h}_{2}}{50}=\frac{-150}{300}$

$\frac{{h}_{2}}{50}=\frac{-1}{2}$

$h_2=\frac{-50}{2}$

$h_2=-25mm$

Thus, the size of the image will be 25 mm, 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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