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Light Reflection and Refraction
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11/08/2021 7:38 pm
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When an object is placed at a distance of 50 cm from a concave spherical mirror, the magnification produced is, -1/2 Where should the object be placed to get a magnification of, -1/5?
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11/08/2021 7:57 pm
Case 1:
u = -50 cm
m = \(-\frac{1}{2}\)
m = -\(\frac{v}{u}\)
-\(\frac{1}{2}\) = -\(\frac{v}{-50}\)
v = -25 cm
We know,
\(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)
⇒ \(\frac{1}{-25}\) + \(\frac{1}{-50}\) = \(\frac{1}{f}\)
⇒ \(\frac{-3}{50}\) = \(\frac{1}{f}\)
⇒ f = \(\frac{50}{3}\) cm
Case 2:
m = -\(\frac{1}{5}\)
f = \(\frac{-50}{3}\) cm
m = -\(\frac{1}{5}\) = -\(\frac{v}{u}\)
v = \(\frac{u}{5}\)
Now,
\(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)
\(\frac{5}{u}\) + \(\frac{1}{u}\) = \(\frac{-3}{50}\)
u = \(\frac{600}{-3}\)
= -100 cm