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An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

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An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

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Given: u = -10 cm, f = 15 cm and v = ?

Applying the formula \(\frac{1}{f}\) = \(\frac{1}{v}\) + \(\frac{1}{u}\), we get

\(\frac{1}{15}\) = \(\frac{1}{v}\) - \(\frac{1}{10}\)

⇒ \(\frac{1}{v}\) = \(\frac{1}{15}\) + \(\frac{1}{10}\)

= \(\frac{2+3}{30}\) = \(\frac{5}{30}\) = \(\frac{1}{6}\)

⇒ v = 6 cm

Hence, the image is formed behind the mirror. It is erect, virtual and smaller than the object.

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