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Rays from the sun converge at a point $15 cm$ in front of a concave mirror. Where should an object be placed so that a real image of size equal to the size of the object is obtained?

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15 cm

in front of the mirror

30 cm

in front of the mirror

Between

15 cm

and

30 cm

in front of the mirror

More than

30 cm

in front of the mirror

answer is B.

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Detailed Solution

The object should be placed

30 cm

in front of the concave mirror to get an image of the same size.
Given,
Focal length,

f = - 15 cm

[Negative: as the mirror is concave]
Magnification,

m = - 1

[Negative: as the image is real]
By using the magnification formula for the mirror,

m = - \frac{v}{u}

\Rightarrow - 1 = - \frac{v}{u}

\Rightarrow v = u

Now by using the mirror formula,

\frac{1}{f} = \frac{1}{v} + \frac{1}{u}

\Rightarrow \frac{1}{- 15} = \frac{1}{u} + \frac{1}{u}

\Rightarrow \frac{1}{- 15} = \frac{2}{u}

\Rightarrow u = - 30 cm

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Rays from the sun converge at a point 15 cm in front of a concave mirror. Where should an object be placed so that a real image of size equal to the size of the object is obtained?

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Rays from the sun converge at a point $15 cm$ in front of a concave mirror. Where should an object be placed so that a real image of size equal to the size of the object is obtained?