Ray Optics

JEE Main level, numerical answer, medium

275 of 295

Question 275 of 295

In a microscope the objective is having focal length fo=2 cmf_{o}=2 \mathrm{~cm} and eye-piece is having focal length fe=4 cmf_{e}=4 \mathrm{~cm}. The tube length is 32 cm. The magnification produced by this microscope for normal adjustment is ____\_\_\_\_.

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Answer: 100

For microscope at normal adjustment:
M=Lfo×DfeM = \frac{L}{f_o} \times \frac{D}{f_e}
where L=32L = 32 cm, fo=2f_o = 2 cm, fe=4f_e = 4 cm, D=25D = 25 cm
M=322×254=16×6.25=100M = \frac{32}{2} \times \frac{25}{4} = 16 \times 6.25 = 100

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