Ray Optics

JEE Main level, single correct, medium

286 of 295

Question 286 of 295

In a compound microscope the focal length of objective lens is 1.2 cm1.2 \mathrm{~cm} and focal length of eye piece is 3.0 cm3.0 \mathrm{~cm}. When object is kept at 1.25 cm1.25 \mathrm{~cm} in front of objective, final image is formed at infinity. Magnifying power of the compound microscope should be:

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Answer: (A) 200

Given : f0=1.2 cm;fe=3.0 cmf_0=1.2 \mathrm{~cm} ; f_e=3.0 \mathrm{~cm} u0=1.25 cm;M∞=u_0=1.25 \mathrm{~cm} ; M_{\infty}= ?
From 1f0=1v0−1u0\frac{1}{f_0}=\frac{1}{v_0}-\frac{1}{u_0}
⇒11.2=1v0−1(−1.25)\Rightarrow \quad \frac{1}{1.2}=\frac{1}{v_0}-\frac{1}{(-1.25)}

⇒1v0=11.2−11.25⇒v0=30 cm\begin{aligned} &\Rightarrow \quad \frac{1}{v_0}=\frac{1}{1.2}-\frac{1}{1.25} \\ &\Rightarrow v_0=30 \mathrm{~cm} \end{aligned}
Magnification at infinity,
M∞=−v0u0×Dfe=301.25×253(∵D=25 cm least distance \begin{aligned} &M_{\infty}=-\frac{v_0}{u_0} \times \frac{D}{f_e} \\ &=\frac{30}{1.25} \times \frac{25}{3} \quad(\because D=25 \mathrm{~cm} \text { least distance } \end{aligned}
of distinct vision)
=200=200
Hence the magnifying power of the
compound microscope is 200

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