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JEE Main 2014 (9 Apr, Shift 1) · Optical Instruments & the Eye

Optical Instruments & the Eye question from JEE Main 2014 (9 Apr, Shift 1)

Q45JEE Main 2014 · 9 APR · SHIFT 1Medium

Optical Instruments & the Eye

The focal lengths of objective lens and eye lens of a Galilean Telescope are respectively 30 cm and 3.0 cm. Telescope produces virtual, erect image of an object situated far away from it at least distance of distinct vision from the eye lens. In this condition, the Magnifying Power of the Galilean Telescope should be:

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Answer: (d)

  1. Galilean eyepiece is concave: fe=−3f_e = -3 cm; final image at v=−25v = -25 cm.
  2. 1ue=1v−1fe=−125+13=2275⇒ue=7522\dfrac{1}{u_e} = \dfrac{1}{v} - \dfrac{1}{f_e} = -\dfrac{1}{25} + \dfrac{1}{3} = \dfrac{22}{75} \Rightarrow u_e = \dfrac{75}{22} cm (virtual object).
  3. M=foue=30×2275=8.8M = \dfrac{f_o}{u_e} = \dfrac{30 \times 22}{75} = 8.8, positive because the image is erect.
  4. So the answer is (d) +8.8+8.8.
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