Answer
64.8k+ views
Hint A concave mirror is a spherical mirror whose reflecting surface is its bent surface. There is a relation between the focal length of a mirror $f$, distance of an object $u$ and distance of the image formed $v$, known as Mirror formula and given by $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ .
Linear magnification of a mirror is the ratio of height of the image formed by the mirror to the height of the object and which is given by $m = \dfrac{{ - v}}{u}$ where $v$ is the distance of the image and $u$ is the distance of the object. If an object is magnified by two times and is inverted then its linear magnification $m = - 2$ .
Complete step by step answer
Let us first know about a concave mirror.
A concave mirror is a spherical mirror whose reflecting surface is its bent surface.
As given in the question that the focal length of a concave mirror, $f = 50{\text{ cm}}$ .
We are asked to find the distance of the object $u$ such that the image formed is two times magnified, real and inverted.
If an object is magnified by two times and is inverted then its linear magnification $m = - 2$ .
Let us discuss the linear magnification of a mirror.
Linear magnification of a mirror is the ratio of height of the image formed by the mirror to the height of the object and which is given by $m = \dfrac{{ - v}}{u}$ where $v$ is the distance of the image and $u$ is the distance of the object.
So, according to the question,
$\dfrac{{ - v}}{u} = - 2$ which simplifies to $v = 2u$
As we know that there is a relation between the focal length of a mirror $f$, distance of an object $u$ and distance of the image formed $v$, known as Mirror formula and given by $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ .
So, substituting the values in the mirror formula we have
$\dfrac{1}{{50}} = \dfrac{1}{{2u}} + \dfrac{1}{u} = \dfrac{3}{{2u}}$
On simplifying we have
$u = 75cm$
Therefore, the object is to be placed at distance $75{\text{ cm}}$ from the mirror so that its image is two times magnified, real and inverted.
Hence, option A is correct
Note Here, we take negative sign for the magnification because the image formed is inverted i.e. below the central axis of the mirror.
Distances of any point are measured along the central axis by convention, as the direction towards the object in mirror is taken as positive and away from the object is taken as negative. Hence the radius of curvature and the focal length are negative for convex mirrors and positive for concave mirrors.
Linear magnification of a mirror is the ratio of height of the image formed by the mirror to the height of the object and which is given by $m = \dfrac{{ - v}}{u}$ where $v$ is the distance of the image and $u$ is the distance of the object. If an object is magnified by two times and is inverted then its linear magnification $m = - 2$ .
Complete step by step answer
Let us first know about a concave mirror.
A concave mirror is a spherical mirror whose reflecting surface is its bent surface.
As given in the question that the focal length of a concave mirror, $f = 50{\text{ cm}}$ .
We are asked to find the distance of the object $u$ such that the image formed is two times magnified, real and inverted.
If an object is magnified by two times and is inverted then its linear magnification $m = - 2$ .
Let us discuss the linear magnification of a mirror.
Linear magnification of a mirror is the ratio of height of the image formed by the mirror to the height of the object and which is given by $m = \dfrac{{ - v}}{u}$ where $v$ is the distance of the image and $u$ is the distance of the object.
So, according to the question,
$\dfrac{{ - v}}{u} = - 2$ which simplifies to $v = 2u$
As we know that there is a relation between the focal length of a mirror $f$, distance of an object $u$ and distance of the image formed $v$, known as Mirror formula and given by $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ .
So, substituting the values in the mirror formula we have
$\dfrac{1}{{50}} = \dfrac{1}{{2u}} + \dfrac{1}{u} = \dfrac{3}{{2u}}$
On simplifying we have
$u = 75cm$
Therefore, the object is to be placed at distance $75{\text{ cm}}$ from the mirror so that its image is two times magnified, real and inverted.
Hence, option A is correct
Note Here, we take negative sign for the magnification because the image formed is inverted i.e. below the central axis of the mirror.
Distances of any point are measured along the central axis by convention, as the direction towards the object in mirror is taken as positive and away from the object is taken as negative. Hence the radius of curvature and the focal length are negative for convex mirrors and positive for concave mirrors.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)