The distance between the lenses in a compound microscope is 15 cm. The focal len

Question

The distance between the lenses in a compound microscope is 15 cm. The focal length of the objective is 1.0 cm. If the microscope is to provide an angular magnification of -81 when used by a person with a normal near point (25 cm from the eye), what must be the focal length of the eyepiece? & Your work in question(s) 8 will also be submitted or saved. One of my telescopes is a 90 mm diameter Meade ETX, a type of Maksutov-Cassegrain. It has an objective with a focal length of 1250 mm. Even though the telescope is a reflecting telescope (uses a mirror instead of a lens as the objective), the formula for the angular magnification is the same. I have two eyepieces, on with a focal length of 26 mm and one with a focal length of 12.4 mm. What is the magnification when the 26-mm eyepiece is used? What is the magnification when the 12.4-mm eyepiece is used? Which eyepiece would be more likely to allow me to see the entire moon in the telescope's field of view? The 26 mm focal length eyepiece. The 12.4 mm focal length eyepiece. Which eyepiece should I use to closely examine a crater on the moon? The 26 mm focal length eyepiece. The 12.4 mm focal length eyepiece.

Explanation / Answer

(1) The focal length of the eyepiece will be given as :

using a formula, we have

M = m0 . Me = - (L / f0) (25 cm / fe)

where, L = distance b/w the lenses = 15 cm

f0 = objective focal length = 1 cm

Me = angular magnification = -81

then, we get

Me = (25 cm / fe)

- (81) = 25 cm / fe

fe = - 0.308 cm

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