
A concave mirror gives an image three times as large as the object when placed at a distance of $20\;{\text{cm}}$ from it. For the image to be real, the focal length should be:
(A) $10\;{\text{cm}}$
(B) $15\;{\text{cm}}$
(C) $20\;{\text{cm}}$
(D) $30\;{\text{cm}}$
Answer
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Hint: If a hollow sphere is cut into parts and the outer surface of the cut part is painted, then it becomes a mirror with its inner surface as the reflecting surface. This mirror type is referred to as a concave mirror.
Formula Used: We will use the following formula to find out the solution of this question
Mirror formula
$\dfrac{1}{{\text{f}}} = \dfrac{1}{{\text{v}}} + \dfrac{1}{{\text{u}}}$
And, magnification formula
${\text{m}} = \dfrac{{ - {\text{v}}}}{{\text{u}}}$
Where
\[m\] is the magnification produced by the concave mirror
\[v\] is the image distance
\[u\] is the object distance
\[f\] is the focal length of the concave mirror
Complete Step-by-Step Solution:
The following information is provided to us in the question:
The magnification produced by the concave mirror, \[m = - 3\]
Object distance, \[u = - 20 cm\]
Now, we will use the magnification formula to find out the image distance
${\text{m}} = \dfrac{{ - {\text{v}}}}{{\text{u}}}$
Upon substituting values, we have
\[ - 3 = \dfrac{{ - v}}{{ - 20}}\]
So, we get Image distance, \[v = - 60 cm\]
Now, we will use the mirror formula
$\dfrac{1}{{\text{f}}} = \dfrac{1}{{\text{v}}} + \dfrac{1}{{\text{u}}}$
On substituting the known values, we get
\[\dfrac{1}{f} = \dfrac{1}{{ - 60}} + \dfrac{1}{{ - 20}}\]
On solving, we get
\[\therefore f = - 15 cm\]
So, the focal length of the concave mirror is \[15 cm\]
Hence, the correct option is (B.)
Note: When light strikes and reflects back from the reflecting surface of the concave mirror, it converges at a point. It is also, therefore, known as a converging mirror. A magnified and virtual image is acquired when the concave mirror is positioned very close to the object.
However, if we increase the distance between the object and the mirror, the image size decreases and a real picture is created. The image that the concave mirror creates can be small or large, or it can be real or virtual.
Formula Used: We will use the following formula to find out the solution of this question
Mirror formula
$\dfrac{1}{{\text{f}}} = \dfrac{1}{{\text{v}}} + \dfrac{1}{{\text{u}}}$
And, magnification formula
${\text{m}} = \dfrac{{ - {\text{v}}}}{{\text{u}}}$
Where
\[m\] is the magnification produced by the concave mirror
\[v\] is the image distance
\[u\] is the object distance
\[f\] is the focal length of the concave mirror
Complete Step-by-Step Solution:
The following information is provided to us in the question:
The magnification produced by the concave mirror, \[m = - 3\]
Object distance, \[u = - 20 cm\]
Now, we will use the magnification formula to find out the image distance
${\text{m}} = \dfrac{{ - {\text{v}}}}{{\text{u}}}$
Upon substituting values, we have
\[ - 3 = \dfrac{{ - v}}{{ - 20}}\]
So, we get Image distance, \[v = - 60 cm\]
Now, we will use the mirror formula
$\dfrac{1}{{\text{f}}} = \dfrac{1}{{\text{v}}} + \dfrac{1}{{\text{u}}}$
On substituting the known values, we get
\[\dfrac{1}{f} = \dfrac{1}{{ - 60}} + \dfrac{1}{{ - 20}}\]
On solving, we get
\[\therefore f = - 15 cm\]
So, the focal length of the concave mirror is \[15 cm\]
Hence, the correct option is (B.)
Note: When light strikes and reflects back from the reflecting surface of the concave mirror, it converges at a point. It is also, therefore, known as a converging mirror. A magnified and virtual image is acquired when the concave mirror is positioned very close to the object.
However, if we increase the distance between the object and the mirror, the image size decreases and a real picture is created. The image that the concave mirror creates can be small or large, or it can be real or virtual.
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