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NEET 2011 (AIPMT, Mains) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from NEET 2011 (AIPMT, Mains)

Q101NEET 2011 · AIPMT · MAINSMedium

Spherical Surfaces & Thin Lenses

A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at a point 15 cm from the lens on the opposite side. If the lens is removed the point where the rays meet will move 5 cm closer to the lens. The focal length of the lens is

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Answer: (d)

  1. Without the lens the rays meet 15−5=1015 - 5 = 10 cm from the lens: this is a virtual object, u=+10u = +10 cm.
  2. With the lens the image is at v=+15v = +15 cm.
  3. 1f=1v−1u=115−110=−130\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{15} - \frac{1}{10} = -\frac{1}{30}
  4. So the answer is (d) f=−30f = -30 cm.
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