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NEET 2026
Exam
NEET
Year
2026
Subject
Physics
Chapter
Chapter

NEET 2026 · Physics · Chapter

An optical system is constructed by placing three thin lenses...

An optical system is constructed by placing three thin lenses coaxially in contact. The focal lengths of the three lenses are ff, f2\dfrac{f}{2}, and −f-f, respectively. If the lateral magnifications produced by these individual lenses are m1m_1, m2m_2, and m3m_3, what is the equivalent power and the total lateral magnification of the system?

Options

  1. A2f\dfrac{2}{f} and m1+m2+m3m_1 + m_2 + m_3
  2. B4f\dfrac{4}{f} and m1m2m3m_1 m_2 m_3
  3. C2f\dfrac{2}{f} and m1m2m3m_1 m_2 m_3
  4. D4f\dfrac{4}{f} and m1+m2+m3m_1 + m_2 + m_3

Correct answer

C. 2f\dfrac{2}{f} and m1m2m3m_1 m_2 m_3

Step-by-step solution

The equivalent power of a combination of thin lenses in contact is the sum of their individual powers.

P=P1+P2+P3P = P_1 + P_2 + P_3

P=1f1+1f2+1f3P = \dfrac{1}{f_1} + \dfrac{1}{f_2} + \dfrac{1}{f_3}

Substituting the given focal lengths f1=ff_1 = f, f2=f2f_2 = \dfrac{f}{2}, and f3=−ff_3 = -f:

P=1f+2f−1f=2fP = \dfrac{1}{f} + \dfrac{2}{f} - \dfrac{1}{f} = \dfrac{2}{f}

For a system of lenses in contact, the total lateral magnification is the product of the lateral magnifications produced by the individual lenses.

M=m1×m2×m3=m1m2m3M = m_1 \times m_2 \times m_3 = m_1 m_2 m_3

Answer: 2f\dfrac{2}{f} and m1m2m3m_1 m_2 m_3
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