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Question

If the focal length of objective and eye lens are $$1.2\ cm$$ and $$3\ cm$$ respectively and the object is put $$1.25\ cm$$ away from the objective lens and the final image is formed at infinity. The magnifying power of the microscope is 


A
150
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B
200
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C
250
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D
400
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Solution

The correct option is A $$200$$
$$m_{\infty}=-\dfrac{v_0}{u_0}\times \dfrac{D}{f_e}$$
From $$\dfrac{1}{f_0}=\dfrac{1}{v_0}-\dfrac{1}{u_0}$$
$$\Rightarrow \dfrac{1}{(+1.2)}=\dfrac{1}{v_0}-\dfrac{1}{(-1.25)}\Rightarrow v_0=30\ cm$$
$$\therefore | m_{ \infty}|=\dfrac{30}{1.25}\times \dfrac{25}{3}=200$$

Physics

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