
How far should one hold an object from a concave mirror of focal length 40 cm so as to get an image twice the size of the object?
Answer
501.6k+ views
Hint:In order to solve this question, we are going to first consider the information about the concave mirror as given in the question, then the image distance in relation to the object distance is found from the magnification. After that the lens equation is used to find the value of the object distance.
Formula used: The magnification is given by the formula,
\[m = \dfrac{{ - v}}{u} = \dfrac{{h'}}{h}\]
Here, \[u\]is the size of the object and\[v\]is the size of the image\[h'\]is the size of image and \[h\]is the size of the object.
Complete step-by-step solution:
In this question, we are given the focal length of the concave mirror equal to:
\[f = 40cm\]
We are given that the size of the image is twice as the size of the object, i.e.,
\[h' = 2h\]
Where, \[h'\]is the size of image
And \[h\]is the size of the object.
We need to find the distance of the object.
Let the size of the object be \[u\]and the size of the image be \[v\]
Here magnification is:
\[m = \dfrac{{h'}}{h} = 2\]
Also, magnification is given by the formula,
\[m = \dfrac{{ - v}}{u}\]
Putting the value of the magnification, we get
\[
m = \dfrac{{ - v}}{u} = 2 \\
\Rightarrow v = - 2u \\
\]
Now the formula for the focal length, i.e., the lens equation,
\[\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}\]
Putting the values, we get
\[
\dfrac{1}{{40}} = \dfrac{1}{u} + \dfrac{1}{{\left( { - 2u} \right)}} \\
\Rightarrow \dfrac{1}{{40}} = \dfrac{1}{u} - \dfrac{1}{{2u}} \\
\Rightarrow \dfrac{1}{{40}} = \dfrac{{2 - 1}}{{2u}} = \dfrac{1}{{2u}} \\
\Rightarrow u = 20cm \\
\]
Thus, one hold an object at a distance of\[20cm\] from a concave mirror of focal length 40 cm so as to get an image twice the size of the object.
Note:It is important to note that the magnification produced by a concave lens is always less than 1.It depends on the distance of the object as well as that of the image and the height of the object as well as that of the image. Lens formula is applicable to both concave and convex lenses which is used to fnd distance of object.
Formula used: The magnification is given by the formula,
\[m = \dfrac{{ - v}}{u} = \dfrac{{h'}}{h}\]
Here, \[u\]is the size of the object and\[v\]is the size of the image\[h'\]is the size of image and \[h\]is the size of the object.
Complete step-by-step solution:
In this question, we are given the focal length of the concave mirror equal to:
\[f = 40cm\]
We are given that the size of the image is twice as the size of the object, i.e.,
\[h' = 2h\]
Where, \[h'\]is the size of image
And \[h\]is the size of the object.
We need to find the distance of the object.
Let the size of the object be \[u\]and the size of the image be \[v\]
Here magnification is:
\[m = \dfrac{{h'}}{h} = 2\]
Also, magnification is given by the formula,
\[m = \dfrac{{ - v}}{u}\]
Putting the value of the magnification, we get
\[
m = \dfrac{{ - v}}{u} = 2 \\
\Rightarrow v = - 2u \\
\]
Now the formula for the focal length, i.e., the lens equation,
\[\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}\]
Putting the values, we get
\[
\dfrac{1}{{40}} = \dfrac{1}{u} + \dfrac{1}{{\left( { - 2u} \right)}} \\
\Rightarrow \dfrac{1}{{40}} = \dfrac{1}{u} - \dfrac{1}{{2u}} \\
\Rightarrow \dfrac{1}{{40}} = \dfrac{{2 - 1}}{{2u}} = \dfrac{1}{{2u}} \\
\Rightarrow u = 20cm \\
\]
Thus, one hold an object at a distance of\[20cm\] from a concave mirror of focal length 40 cm so as to get an image twice the size of the object.
Note:It is important to note that the magnification produced by a concave lens is always less than 1.It depends on the distance of the object as well as that of the image and the height of the object as well as that of the image. Lens formula is applicable to both concave and convex lenses which is used to fnd distance of object.
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