Answer
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Hint: Magnification of a convex mirror is defined as the ratio of image distance to the object distance. Mirror formula says that the reciprocal of the focal length of a convex mirror is equal to the sum of reciprocals of image distance and object distance. Combining both these concepts, we can determine the distance of the object from the given convex mirror.
Formula used:
$1)M=-\dfrac{v}{u}$
$2)\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$
Complete step-by-step solution
Magnification refers to the process of enlargement of an object. The magnification of a convex mirror refers to the change in the size of the object when viewed using a convex mirror. Mathematically, magnification of a convex mirror is given by
$M=-\dfrac{v}{u}$
where
$M$ is the magnification of a convex mirror
$v$ is the distance of the image of an object from a convex mirror
$u$ is the distance of the object from the convex mirror
Let this be equation 1.
At the same time, the mirror formula suggests that the reciprocal of the focal length of a convex mirror is equal to the sum of reciprocals of object distance as well as image distance from the convex mirror. Mathematically, the mirror formula is given by
$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$
where
$f$ is the focal length of a convex mirror
$u$ is the distance of an object from the convex mirror
$v$ is the distance of the image of the object from the convex mirror
Let this be equation 2.
Combining both equation 1 and equation 2, we have
$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}\Rightarrow \dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{-(Mu)}\Rightarrow\dfrac{1}{f}=\dfrac{1}{u}\left( 1-\dfrac{1}{M} \right)$
where
$f$ is the focal length of a convex mirror
$u$ is the object distance from the convex mirror
$M$ is the magnification of the convex mirror
Let this be equation 3.
Coming to our question, we are provided with a convex mirror of focal length $20cm$, which when used to magnify an object gives an image, whose size is a quarter of the object. We are required to determine the distance of the object from the mirror.
Clearly, we are given that
$M=\dfrac{1}{4}$, is the magnification of the convex mirror
and
$f=20cm=0.2m$, is the focal length of the convex mirror
Substituting these values in equation 3, we have
$\dfrac{1}{f}=\dfrac{1}{u}\left( 1-\dfrac{1}{M} \right)\Rightarrow \dfrac{1}{0.2m}=\dfrac{1}{u}\left( 1-\dfrac{1}{\left( \dfrac{1}{4} \right)} \right)\Rightarrow \dfrac{1}{0.2m}=\dfrac{1}{u}\left( 1-4 \right)=-\dfrac{3}{u}\Rightarrow u=-0.6m=-60cm$
Therefore, the distance of the object from the given convex mirror is equal to $60cm$.
Note: Mirror formula as well as formula for magnification follows sign convention. A negative sign in the formula for magnification (equation 1) of a convex mirror suggests that the image formed has less size than the size of the object. Similarly, the negative sign of object distance in the final answer suggests that the object is placed on the side of the mirror, opposite to the side where the focal length of the convex mirror falls. Students need to be aware that magnification of a convex mirror is also given by
$M=\dfrac{h}{{{h}_{0}}}$
where
$h$ is the height or size of the image of an object
${{h}_{0}}$ is the height or size of the object
Formula used:
$1)M=-\dfrac{v}{u}$
$2)\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$
Complete step-by-step solution
Magnification refers to the process of enlargement of an object. The magnification of a convex mirror refers to the change in the size of the object when viewed using a convex mirror. Mathematically, magnification of a convex mirror is given by
$M=-\dfrac{v}{u}$
where
$M$ is the magnification of a convex mirror
$v$ is the distance of the image of an object from a convex mirror
$u$ is the distance of the object from the convex mirror
Let this be equation 1.
At the same time, the mirror formula suggests that the reciprocal of the focal length of a convex mirror is equal to the sum of reciprocals of object distance as well as image distance from the convex mirror. Mathematically, the mirror formula is given by
$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$
where
$f$ is the focal length of a convex mirror
$u$ is the distance of an object from the convex mirror
$v$ is the distance of the image of the object from the convex mirror
Let this be equation 2.
Combining both equation 1 and equation 2, we have
$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}\Rightarrow \dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{-(Mu)}\Rightarrow\dfrac{1}{f}=\dfrac{1}{u}\left( 1-\dfrac{1}{M} \right)$
where
$f$ is the focal length of a convex mirror
$u$ is the object distance from the convex mirror
$M$ is the magnification of the convex mirror
Let this be equation 3.
Coming to our question, we are provided with a convex mirror of focal length $20cm$, which when used to magnify an object gives an image, whose size is a quarter of the object. We are required to determine the distance of the object from the mirror.
Clearly, we are given that
$M=\dfrac{1}{4}$, is the magnification of the convex mirror
and
$f=20cm=0.2m$, is the focal length of the convex mirror
Substituting these values in equation 3, we have
$\dfrac{1}{f}=\dfrac{1}{u}\left( 1-\dfrac{1}{M} \right)\Rightarrow \dfrac{1}{0.2m}=\dfrac{1}{u}\left( 1-\dfrac{1}{\left( \dfrac{1}{4} \right)} \right)\Rightarrow \dfrac{1}{0.2m}=\dfrac{1}{u}\left( 1-4 \right)=-\dfrac{3}{u}\Rightarrow u=-0.6m=-60cm$
Therefore, the distance of the object from the given convex mirror is equal to $60cm$.
Note: Mirror formula as well as formula for magnification follows sign convention. A negative sign in the formula for magnification (equation 1) of a convex mirror suggests that the image formed has less size than the size of the object. Similarly, the negative sign of object distance in the final answer suggests that the object is placed on the side of the mirror, opposite to the side where the focal length of the convex mirror falls. Students need to be aware that magnification of a convex mirror is also given by
$M=\dfrac{h}{{{h}_{0}}}$
where
$h$ is the height or size of the image of an object
${{h}_{0}}$ is the height or size of the object
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