## Related questions with answers

A ball is positioned 22 cm in front of a spherical mirror and forms a virtual image. If the spherical mirror is replaced with a plane mirror, the image appears 12 cm closer to the mirror. What kind of spherical mirror was used?

Solutions

VerifiedMirror Equation :

$\dfrac{1}{f}=\dfrac{1}{d_o}+\dfrac{1}{d_i}$

Given that : $d_o = 22$

For Images produced by plane mirrors, object and Image are at the same distance from the mirror

That means that the Image is now 22cm behind the mirror. But before the plane mirror was used, the image was $22+12 = 34$ cm behind the mirror

Therefore $d_i = -34$

$\dfrac{1}{f} = \dfrac{1}{22}+\dfrac{1}{-34}$

$\dfrac{1}{f} = \dfrac{1}{22}\times \dfrac{34}{34}-\dfrac{1}{34}\times \dfrac{22}{22}$

$\dfrac{1}{f} = \dfrac{34}{34\times22}-\dfrac{22}{34\times22}$

$\dfrac{1}{f} = \dfrac{12}{34\times22}$

Take reciprocal of both sides

${f} = \dfrac{34\times22}{12}\approx62.3\text{ cm}$

The positive sign means that the mirror was concave

In this problem, we need to find the type of mirror (concave or convex). We can do that by observing the focal length of the mirror. We know where the ball is positioned, and the mirror forms the virtual image, $d_{i}<0$. Also, we know that if we replace the spherical mirror with a plane the image is closer for $12 \hspace{0.5mm} \mathrm{cm}$.

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