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JEE Main 2024 (1 Feb, Shift 1) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from JEE Main 2024 (1 Feb, Shift 1)

Q46JEE Main 2024 · 1 FEB · SHIFT 1Medium

Spherical Surfaces & Thin Lenses

The distance between object and its 3 times magnified virtual image as produced by a convex lens is 20 cm20\text{ cm}. The focal length of the lens used is __________ cm\text{cm}.

Numerical answer type. Work it out, then open the answer.

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Answer: 15

  1. Virtual image on the same side: v=3uv = 3u and ∣v∣−∣u∣=20|v| - |u| = 20 cm.
  2. 2∣u∣=20⇒u=−102|u| = 20 \Rightarrow u = -10 cm, v=−30v = -30 cm
  3. 1f=1v−1u=−130+110=230\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = -\frac{1}{30} + \frac{1}{10} = \frac{2}{30}
  4. f=15f = 15 cm
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