Question:
In a compound microscope the focal length of objective lens is \(1.2\ \mathrm{cm}\) and focal length of eye piece is \(3.0\ \mathrm{cm}\). When object is kept at \(1.25\ \mathrm{cm}\) in front of objective, final image is formed at infinity. Magnifying power of the compound microscope should be:
In a compound microscope the focal length of objective lens is \(1.2\ \mathrm{cm}\) and focal length of eye piece is \(3.0\ \mathrm{cm}\). When object is kept at \(1.25\ \mathrm{cm}\) in front of objective, final image is formed at infinity. Magnifying power of the compound microscope should be:
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Find the objective image distance from the lens formula, then use \(M=m_o\times m_e\) with \(m_e=D/f_e\) for image at infinity.
Updated On: Nov 11, 2026
- 200
- 100
- 400
- 150
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The Correct Option is A
Solution and Explanation
Objective: \(u=-1.25\) cm, \(f_o=1.2\) cm.
\(\frac{1}{v}=\frac{1}{f_o}+\frac{1}{u}=\frac{1}{1.2}-\frac{1}{1.25}=\frac{0.05}{1.5}\), so \(v=30\) cm.
\(m_o=\frac{v}{u}=\frac{30}{1.25}=24\) in magnitude.
Final image at infinity: \(m_e=\frac{D}{f_e}=\frac{25}{3}\).
\(M=24\times\frac{25}{3}=200\).
Answer: (1) 200
\(\frac{1}{v}=\frac{1}{f_o}+\frac{1}{u}=\frac{1}{1.2}-\frac{1}{1.25}=\frac{0.05}{1.5}\), so \(v=30\) cm.
\(m_o=\frac{v}{u}=\frac{30}{1.25}=24\) in magnitude.
Final image at infinity: \(m_e=\frac{D}{f_e}=\frac{25}{3}\).
\(M=24\times\frac{25}{3}=200\).
Answer: (1) 200
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