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JEE Main 2026 (23 Jan, Shift 2) · Spherical Surfaces & Thin Lenses

Spherical Surfaces & Thin Lenses question from JEE Main 2026 (23 Jan, Shift 2)

Q18JEE Main 2026 · 23 JAN · SHIFT 2Medium

Spherical Surfaces & Thin Lenses

The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions 8 cm and 24 cm from the lens. The focal length of the lens is ______ cm.

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

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Answer: 16

  1. Equal size: one image real, one virtual, so m1=−m2m_1 = -m_2 with m=ff+um = \frac{f}{f + u}.
  2. ff−8=−ff−24\frac{f}{f - 8} = -\frac{f}{f - 24}
  3. f−8=24−f⇒2f=32f - 8 = 24 - f \Rightarrow 2f = 32
  4. f=16f = 16 cm
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