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JEE Main 2015 (10 Apr, Shift 1) · Reflection & Mirrors

Reflection & Mirrors question from JEE Main 2015 (10 Apr, Shift 1)

Q48JEE Main 2015 · 10 APR · SHIFT 1Medium

Reflection & Mirrors

You are asked to design a shaving mirror assuming that a person keeps it 10 cm from his face and views the magnified image of the face at the closest comfortable distance of 25 cm. The radius of curvature of the mirror would then be:

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Answer: (c)

  1. u=−10 cmu = -10\text{ cm}. The virtual image is behind the mirror and 25 cm from the face, so v=+15 cmv = +15\text{ cm}.
  2. 1f=1v+1u=115−110=−130⇒f=−30 cm\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{15} - \frac{1}{10} = -\frac{1}{30} \Rightarrow f = -30\text{ cm}
  3. ∣R∣=2∣f∣=60 cm|R| = 2|f| = 60\text{ cm} (concave mirror).
  4. So the answer is (c) 60 cm.
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