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MHT CET 2025 (20 Apr, Shift 1) · Optical Instruments & the Eye

Optical Instruments & the Eye question from MHT CET 2025 (20 Apr, Shift 1)

Q8MHT CET 2025 · 20 APR · SHIFT 1Medium

Optical Instruments & the Eye

The length of the compound microscope is 15 cm. The magnifying power for relaxed eye is 25. If the focal length of eye lens is 6 cm then the object distance for objective lens will be

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Answer: (b)

  1. Relaxed eye: eyepiece object at fef_e, so objective image distance vo=L−fe=15−6=9v_o = L - f_e = 15 - 6 = 9 cm.
  2. M=vouo⋅Dfe=9uo⋅256M = \dfrac{v_o}{u_o} \cdot \dfrac{D}{f_e} = \dfrac{9}{u_o} \cdot \dfrac{25}{6}.
  3. 25=9uo⋅256⇒uo=1.525 = \dfrac{9}{u_o} \cdot \dfrac{25}{6} \Rightarrow u_o = 1.5 cm.
  4. So the answer is (b) 1.5 cm.
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