Answer
Verified
44.1k+ views
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, $m = \dfrac{{ - v}}{u} = \dfrac{{{h_1}}}{{{h_2}}}$
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=$\dfrac{1}{2}$for first case:
$\dfrac{1}{2} = \dfrac{{ - v}}{{ - 60}} = v = 30cm$
Now applying the mirror equation:
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$
After putting the value of u and v in the equation:
$
\dfrac{1}{{30}} + \dfrac{1}{{( - 60)}} = \dfrac{1}{f} \\
\Rightarrow \dfrac{1}{f} = \dfrac{1}{{30}} - \dfrac{1}{{60}} \\
\Rightarrow \dfrac{1}{f} = \dfrac{{2 - 1}}{{60}} \\
\Rightarrow f = 60cm \\
$
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 $\dfrac{1}{3}$
So, $
\dfrac{1}{3} = \dfrac{{ - v}}{u} \\
\Rightarrow v = \dfrac{{ - u}}{3} \\
$
Now putting the value of v in the mirror equation:
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$
$
\dfrac{1}{{\dfrac{{ - u}}{3}}} + \dfrac{1}{u} = \dfrac{1}{{60}} \\
\Rightarrow \dfrac{{ - 3}}{u} + \dfrac{1}{u} = \dfrac{1}{{60}} \\
\Rightarrow \dfrac{{ - 3 + 1}}{u} = \dfrac{1}{{60}} \\
\Rightarrow u = - 120cm \\
$
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, $m = \dfrac{{ - v}}{u} = \dfrac{{{h_1}}}{{{h_2}}}$
It is also calculated by the ratio of height of image to the height of the object.
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, $m = \dfrac{{ - v}}{u} = \dfrac{{{h_1}}}{{{h_2}}}$
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=$\dfrac{1}{2}$for first case:
$\dfrac{1}{2} = \dfrac{{ - v}}{{ - 60}} = v = 30cm$
Now applying the mirror equation:
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$
After putting the value of u and v in the equation:
$
\dfrac{1}{{30}} + \dfrac{1}{{( - 60)}} = \dfrac{1}{f} \\
\Rightarrow \dfrac{1}{f} = \dfrac{1}{{30}} - \dfrac{1}{{60}} \\
\Rightarrow \dfrac{1}{f} = \dfrac{{2 - 1}}{{60}} \\
\Rightarrow f = 60cm \\
$
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 $\dfrac{1}{3}$
So, $
\dfrac{1}{3} = \dfrac{{ - v}}{u} \\
\Rightarrow v = \dfrac{{ - u}}{3} \\
$
Now putting the value of v in the mirror equation:
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$
$
\dfrac{1}{{\dfrac{{ - u}}{3}}} + \dfrac{1}{u} = \dfrac{1}{{60}} \\
\Rightarrow \dfrac{{ - 3}}{u} + \dfrac{1}{u} = \dfrac{1}{{60}} \\
\Rightarrow \dfrac{{ - 3 + 1}}{u} = \dfrac{1}{{60}} \\
\Rightarrow u = - 120cm \\
$
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, $m = \dfrac{{ - v}}{u} = \dfrac{{{h_1}}}{{{h_2}}}$
It is also calculated by the ratio of height of image to the height of the object.
Recently Updated Pages
To get a maximum current in an external resistance class 1 physics JEE_Main
f a body travels with constant acceleration which of class 1 physics JEE_Main
A hollow sphere of mass M and radius R is rotating class 1 physics JEE_Main
If the beams of electrons and protons move parallel class 1 physics JEE_Main
Two radioactive nuclei P and Q in a given sample decay class 1 physics JEE_Main
silver wire has diameter 04mm and resistivity 16 times class 12 physics JEE_Main
Other Pages
Explain the construction and working of a GeigerMuller class 12 physics JEE_Main
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
The highest oxidation state shown by element with atomic class 11 chemistry JEE_Main