A convex mirror used for rear view on an automobile has a radius of curvature of 3m. If a bus is located at 5m from this mirror, find magnification?
A. +0.23
B. -0.23
C. +0.45
D. -0.45
Answer
629.4k+ views
Hint: Use the formula for the magnification in case of a spherical mirror. Also use the mirror formula to find an expression for magnification in terms of the object position and the focal length of the mirror.
Formula used:
$m=-\dfrac{v}{u}$
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$
where $v$ and $u$ are the positions of the image and the object according to the sign convection.
Complete step by step solution:
The magnification due to a spherical mirror is given as $m=-\dfrac{v}{u}$, where v and u are the positions of the image and the object according to the sign convection.
$v=-mu$.
The relation between the position of the object, its image and focal length of the mirror is given as $\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$, where f is the focal length of the mirror.
Substitute the values of the v in the above equation.
$\dfrac{1}{f}=\dfrac{1}{-mu}+\dfrac{1}{u}$ ….. (i)
It is given that the radius of curvature is 3m. The focal length of a spherical mirror is given as $f=\dfrac{R}{2}$, where R is the radius of curvature of the mirror. According to the sign convection, the radius of curvature of a convex mirror is positive. This means that $R=3m$.
$f=\dfrac{3}{2}=1.5m$
It is said that the bus in front of the mirror is at a distance of 5m. This means that $u=-5m$ .
Substitute the values of u and f in (i).
$\dfrac{1}{1.5}=\dfrac{1}{-m(-5)}+\dfrac{1}{-5}$
$\Rightarrow \dfrac{1}{1.5}+\dfrac{1}{5}=\dfrac{1}{5m}$
$\Rightarrow \dfrac{2}{3}+\dfrac{1}{5}=\dfrac{1}{5m}$
$\Rightarrow \dfrac{13}{15}=\dfrac{1}{5m}$
$\therefore m=\dfrac{3}{13}=0.23$.
Therefore, the magnification in this case is equal to +0.23.
Hence, the correct option is A.
Note: The magnification is positive when the image formed is erect and it is negative when the image formed is inverted. Therefore, the image formed by the convex mirror is erect. It is also virtual and formed inside the mirror.Magnification is the increase in the image size produced by spherical mirrors with respect to the object size. It is the ratio of the height of the image to the height of the object and is denoted as m.
Formula used:
$m=-\dfrac{v}{u}$
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$
where $v$ and $u$ are the positions of the image and the object according to the sign convection.
Complete step by step solution:
The magnification due to a spherical mirror is given as $m=-\dfrac{v}{u}$, where v and u are the positions of the image and the object according to the sign convection.
$v=-mu$.
The relation between the position of the object, its image and focal length of the mirror is given as $\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$, where f is the focal length of the mirror.
Substitute the values of the v in the above equation.
$\dfrac{1}{f}=\dfrac{1}{-mu}+\dfrac{1}{u}$ ….. (i)
It is given that the radius of curvature is 3m. The focal length of a spherical mirror is given as $f=\dfrac{R}{2}$, where R is the radius of curvature of the mirror. According to the sign convection, the radius of curvature of a convex mirror is positive. This means that $R=3m$.
$f=\dfrac{3}{2}=1.5m$
It is said that the bus in front of the mirror is at a distance of 5m. This means that $u=-5m$ .
Substitute the values of u and f in (i).
$\dfrac{1}{1.5}=\dfrac{1}{-m(-5)}+\dfrac{1}{-5}$
$\Rightarrow \dfrac{1}{1.5}+\dfrac{1}{5}=\dfrac{1}{5m}$
$\Rightarrow \dfrac{2}{3}+\dfrac{1}{5}=\dfrac{1}{5m}$
$\Rightarrow \dfrac{13}{15}=\dfrac{1}{5m}$
$\therefore m=\dfrac{3}{13}=0.23$.
Therefore, the magnification in this case is equal to +0.23.
Hence, the correct option is A.
Note: The magnification is positive when the image formed is erect and it is negative when the image formed is inverted. Therefore, the image formed by the convex mirror is erect. It is also virtual and formed inside the mirror.Magnification is the increase in the image size produced by spherical mirrors with respect to the object size. It is the ratio of the height of the image to the height of the object and is denoted as m.
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