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Naya [18.7K]
3 years ago
12

A very delicate sample is placed 0.150 cm from the objective lens of a microscope. The focal length of the objective is 0.140 cm

, and that of the eyepiece is 1.0 cm. The near-point distance of the person using the microscope is 25.0 cm. The magnitude of the final magnification of the microscope is closes to:
Physics
1 answer:
iVinArrow [24]3 years ago
4 0

Answer:

m = 14*26 = 364

Explanation:

overall magnification is given as m

m = m_{o}* m_{e}

mo magnification of objective lens

me magnification of EYE lens

where mo is given as

m_{o} = \frac{v_{o}}{-u _{0}}

and me as

m_{e} = 1+\frac{D}{f_{e}}

d is distant of distinct vision = 25.0 cm for normal eye

fe =  focal length of eye piece

focal length of objective lense is 0.140 cm

we know that

\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}

\frac{1}{v_{0}} = \frac{1}{u_{0}} + \frac{1}{f_{0}}

\frac{1}{v_{0}} = \frac{1}{0.150} + \frac{1}{0.14}

\frac{1}{v_{o}} = 2.1 cm

m_{o} = \frac{2.1}{0.150} = 14

m_{e} = 1+\frac{25}{1}

m_{e} =26

m = m_{o}* m_{e}

m = 14*26 = 364

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