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JEE Main 2026 (22 Jan, Shift 1) · Combination of Lenses

Combination of Lenses question from JEE Main 2026 (22 Jan, Shift 1)

Q4JEE Main 2026 · 22 JAN · SHIFT 1Medium

Combination of Lenses

A thin convex lens of focal length 5 cm and a thin concave lens of focal length 4 cm are combined together (without any gap) and this combination has magnification m1m_1 when an object is placed 10 cm before the convex lens. Keeping the positions of convex lens and object undisturbed a gap of 1 cm is introduced between the lenses by moving the concave lens away, which lead to a change in magnification of total lens system to m2m_2. The value of ∣m1m2∣\left| \frac{m_1}{m_2} \right| is ______.

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Answer: (b)

  1. In contact: 1F=15−14=−120\frac{1}{F} = \frac{1}{5} - \frac{1}{4} = -\frac{1}{20}; u=−10u = -10: 1v=−120−110⇒v=−203\frac{1}{v} = -\frac{1}{20} - \frac{1}{10} \Rightarrow v = -\frac{20}{3}, m1=vu=23m_1 = \frac{v}{u} = \frac{2}{3}.
  2. With gap, convex lens: u=−10u = -10, f=5⇒v=10 cmf = 5 \Rightarrow v = 10\text{ cm}, m=−1m = -1.
  3. Concave lens is 11 cm away, so u=−1 cmu = -1\text{ cm}: 1v=−14−1⇒v=−0.8 cm\frac{1}{v} = -\frac{1}{4} - 1 \Rightarrow v = -0.8\text{ cm}, m=0.8m = 0.8.
  4. m2=−1×0.8=−0.8m_2 = -1 \times 0.8 = -0.8, so ∣m1m2∣=2/30.8=56\left|\frac{m_1}{m_2}\right| = \frac{2/3}{0.8} = \frac{5}{6}
  5. So the answer is (b) 56\frac{5}{6}.
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