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Given: Assume the near point (of the eye) is 25 cm . The distance between eyepiece and objective lens in a certain compound microscope is l = 17.6 cm . The focal length of the eyepiece is fe = 2.57 cm , and that of the objective lens is fo = 0.3 cm . What is the overall magnification of the microscope? Caution: a negative quantity this is. Use the approximation l − fe ≈ l and object distance do is approximately the focal length fo.

Respuesta :

The overall magnification of the microscope is - 570.687.

A microscope is a laboratory tool used to have a look at items which can be too small to be seen with the aid of the bare eye. Microscopy is the technology of investigating small objects and structures through the use of a microscope. The microscopic approach is invisible to the attention unless aided by using a microscope.

The focal length of a microscope objective is generally between 2 mm and 40 mm. but, that parameter is regularly considered as less important, on account that magnification and numerical aperture are sufficient for quantifying the critical performance in a microscope.

Calculation:-

M = -(L/f) (near point/Fe)

M = - (0.176/0.003)(0.25/0.0257)

M = - 570.687

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