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Light Reflection and Refraction
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11/08/2021 2:26 pm
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An object of 5.0 cm size is placed at a distance of 20.0 cm from a converging mirror of focal length 15.0 cm. At what distance from the mirror should a screen be placed to get the sharp image? Also, calculate the size of the image.
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11/08/2021 2:31 pm
\(h_1\) = 50 cm, u = 20 cm, f = -15 cm, v = ?, \(h_2\) = ?
We know that
\(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)
\(\frac{1}{v}\) + \(\frac{1}{-20}\) = \(\frac{1}{-15}\)
\(\frac{1}{v}\) = \(\frac{1}{20}\) - \(\frac{1}{15}\)
\(\frac{1}{v}\) = \(\frac{-5}{300}\)
v = -60 cm
The screen should be placed 60 cm in front of the mirror.
And
m = \(\frac{h_2}{h_1}\) = -\(\frac{v}{u}\)
\(\frac{h_2}{5}\) = -\(\frac{-60}{-20}\)
\({h_2}\) = -15 cm
Height of image = 15 cm