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anzhelika [568]
3 years ago
5

The distance between the eyepiece and the objective lens in a certain compound microscope is 27.7 cm. The focal length of the ob

jective is 0.490 cm, and that of the eyepiece is 1.60 cm. Find the overall magnification of the microscope. (The near point of the eye is 25 cm. Assume that the object is placed at the focal point of the objective lens, and one places the eyepiece at the near point of the eye.)
Physics
2 answers:
jonny [76]3 years ago
6 0

Answer:

Magnification, m = -883.29

Explanation:

Given that,

The distance between the eyepiece and the objective lens in a certain compound microscope is 27.7 cm, L = 27.7 cm

The focal length of the objective is 0.490 cm, f_o=0.49\ cm

The focal length of the eye piece is 0.490 cm, f_e=1.6\ cm

The overall magnification of the microscope is given by the following relation :

m=-\dfrac{L}{f_o}(\dfrac{25\ cm}{f_e})

Putting all the values in above formula :

m=-\dfrac{27.7}{0.49}(\dfrac{25\ cm}{1.6})\\\\m=-883.29

So, the overall magnification of the microscope is (-883.29). Hence, this is the required solution.

AnnZ [28]3 years ago
5 0

Answer:

Ocular is the correct answer to this

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