Question

when an object Is placed at a distance of 60 cm. from a convex mirror, the magnification produced Is ${\displaystyle \frac{1}{2}}$. Where should the object be placed to get a magnification of ${\displaystyle \frac{1}{3}}$.

Answer 1

Hint: for solving this question we should have to be familiar with the term magnification.
After applying the definition of magnification firstly we will get the value of the image when the object is placed at 60 cm from the mirror. Then we will find the focus of the mirror. After finding the focus of the mirror we will make a relation between object distance and image distance from the mirror and by the equation we will solve this.

Complete Step by step process
Firstly we all know that magnification is defined as the ratio of height of image and height of the object.
Mathematically, $m={\displaystyle \frac{-v}{u}}={\displaystyle \frac{h_1}{h_2}}$
Where, m is magnification
$v$ is distance of image and the mirror
$u$ is the distance between object and mirror
$\therefore$we have given $u$=-60
And m=${\displaystyle \frac{1}{2}}$for first case:
${\displaystyle \frac{1}{2}}={\displaystyle \frac{-v}{-60}}=v=30cm$
Now applying the mirror equation:
${\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$
After putting the value of u and v in the equation:
$$\begin{gathered} {\displaystyle \frac{1}{30}}+{\displaystyle \frac{1}{(-60)}}={\displaystyle \frac{1}{f}} \\ \Rightarrow {\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{30}}-{\displaystyle \frac{1}{60}} \\ \Rightarrow {\displaystyle \frac{1}{f}}={\displaystyle \frac{2-1}{60}} \\ \Rightarrow f=60cm \end{gathered}$$
Now we have found that the focal length of the mirror is 60 cm.
Now again applying the second case of magnification in the mirror.
We have given magnification is ${\displaystyle \frac{1}{3}}$
So, $$\begin{gathered} {\displaystyle \frac{1}{3}}={\displaystyle \frac{-v}{u}} \\ \Rightarrow v={\displaystyle \frac{-u}{3}} \end{gathered}$$
Now putting the value of v in the mirror equation:
${\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$
$$\begin{gathered} {\displaystyle \frac{1}{{\displaystyle \frac{-u}{3}}}}+{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{60}} \\ \Rightarrow {\displaystyle \frac{-3}{u}}+{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{60}} \\ \Rightarrow {\displaystyle \frac{-3+1}{u}}={\displaystyle \frac{1}{60}} \\ \Rightarrow u=-120cm \end{gathered}$$
so to get the magnification of 1/3 we should place the object at a distance of 120 cm from the mirror.

Therefore, the correct answer will be 120 cm.

Note
Magnification is defined as the ratio of distance between image to mirror and distance between object to mirror multiplied by minus sign.
Mathematically, $m={\displaystyle \frac{-v}{u}}={\displaystyle \frac{h_1}{h_2}}$
It is also calculated by the ratio of height of image to the height of the object.