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A concave mirror produces a real image of half the size of an object placed at 60 cm in front of it. Where should the object be placed to obtain a virtual image of double the size of the object in front of the mirror.

Last updated date: 13th Jun 2024
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Hint: We know that concave mirrors produce real images for the object distance greater than the focal length of the concave mirror. We are given that an object placed in front of the mirror at a distance of 60 cm forms a real image with half its size.
We know the magnification of a concave mirror can be given as the ratio of the image distance to the object distance. It is given as –
We are given the magnification as half and the object distance as 60 cm from the center of the mirror.
  & m=\dfrac{v}{u} \\
 & \Rightarrow -(-\dfrac{1}{2})=\dfrac{v}{-60} \\
 & \therefore v=-30cm \\
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The focal length of the concave mirror can be found as –
  & \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u} \\
 & \Rightarrow \dfrac{1}{f}=\dfrac{1}{-30}+\dfrac{1}{-60} \\
 & \Rightarrow \dfrac{1}{f}=\dfrac{1}{-20}cm \\
 & \therefore f=-20cm \\
We know from the mirror formula that –
Substituting this in the formula for magnification gives us the relation that –
  & m'=-\dfrac{v}{u} \\
 & \Rightarrow m'=-\dfrac{\dfrac{u'f}{u'-f}}{u'} \\
 & \therefore m'=-\dfrac{f}{u'-f} \\
We can substitute the known values of the magnification and the focal length to get the object distance which can create an image which is virtual and double the size of the object. It can be substituted as –
  & m'=-\dfrac{f}{u'-f} \\
 & \Rightarrow 2=-\dfrac{-20}{u'+20} \\
 & \Rightarrow 2u'+40=20 \\
 & \therefore u'=-10cm \\
The object should be placed 10 cm in front of the concave mirror to obtain an image which is virtual and double the size of the object.

The sign conventions should be used appropriately to obtain correct solutions for the physical parameters involved in finding the object distance or any unknown distance. We should take care not to substitute signs more than once for the same quantity.