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Geometrical Optics question

2011 · Q89
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Geometrical Optics question

2011 · Q89

NEETPhysicsGeometrical OpticsMCQ+4 / −1
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
  1. A
    5 cm
  2. B
    −-− 10 cm
  3. C
    20 cm
  4. D
    −-− 30 cm
View written solutionFree

Correct answer: D

Here, v = +15 cm, u = +(15 – 5) = +10 cm

According to lens formula

1v−1u=1f{1 \over v} - {1 \over u} = {1 \over f}v1​−u1​=f1​

⇒\Rightarrow⇒ 115−110=1f{1 \over {15}} - {1 \over {10}} = {1 \over f}151​−101​=f1​

⇒\Rightarrow⇒ f = -30 cm

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