Question

An object is placed at a distance of 16 cm from the pole of a concave mirror of focal length 10 cm. What will
be the nature of the image formed? Justify your answer.​

Answer 1

The distance of object from the mirror,

• $\sf{u=-16\ cm}$

The focal length of the mirror,

• $\sf{f=-10\ cm}$

We have, the mirror formula,

$\sf{\longrightarrow \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}}$

Then, distance of image from the mirror is given by,

$\sf{\longrightarrow\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}}$

$\sf{\longrightarrow\dfrac{1}{v}=\dfrac{1}{-10}-\dfrac{1}{-16}}$

$\sf{\longrightarrow\dfrac{1}{v}=\dfrac{1}{16}-\dfrac{1}{10}}$

$\sf{\longrightarrow\dfrac{1}{v}=-\dfrac{3}{80}}$

$\sf{\longrightarrow v=-\dfrac{80}{3}\ cm}$

Then magnification,

$\sf{\longrightarrow m=-\dfrac{v}{u}}$

$\sf{\longrightarrow m=-\dfrac{\left(-\dfrac{80}{3}\right)}{-16}}$

$\sf{\longrightarrow m=-\dfrac{5}{3},\quad |m|=\dfrac{5}{3}}$

• Here $\sf{v<0.}$ So the image is real.

• Here $\sf{m<0.}$ So the image is inverted.

• Here $\sf{|m|>1.}$ So the image is magnified.

Hence the image is real, inverted and magnified.

Answer 2

Given :-

An object is placed at a distance of 16 cm from the pole of a concave mirror of focal length 10 cm.

To Find :-

Nature of the image formed

Solution :-

We know that

$\sf \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$

According to the question

$\sf We \; have \begin{cases} \rm Object \; Distance_{u} = \frak{-16 \; cm} \\ \rm Image \; Distance_{v} = \frak{v} \\ \rm Focal \; Length_{f} = \frak{-10 \; cm}\end{cases}$

By putting values

$\sf \dfrac{1}{-10} = \dfrac{1}{-16} + \dfrac{1}{v}$

$\sf \dfrac{1}{10} = \dfrac{1}{16} + \dfrac{1}{v}$

$\sf \dfrac{1}{16} - \dfrac{1}{10} = \dfrac{1}{v}$

$\sf \dfrac{5 - 8}{80} = \dfrac{1}{v}$

$\sf\dfrac{-3}{80} = \dfrac{1}{v}$

$\sf \dfrac{-80}{3} = v$

$\sf -26.66 = v$

Finding magnification

$\sf M= \dfrac{-v}{u}$

$\sf M = \dfrac{-(-26.66)}{-16}$

$\sf M = \dfrac{26.66}{16}$

$\sf M = 1.66$

Now

Nature - Real, larger than object, magnification