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A 7.5x binocular produces an angular magnification of negative 7.50, acting like a telescope.
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Mirrors are used to make the image upright.
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If the binoculars have objective lenses with a 75.0-centimeter focal length, what is the focal length of the eyepiece lenses?
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Weβre told in this statement that the angular magnification of the binoculars that act like a telescope is negative 7.50 and that mirrors are used to make the image upright; weβll call this magnification capital π.
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Weβre also told that, in this set of lenses, the objective lens has a 75.0-centimeter focal length, what weβll call π sub π.
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We want to solve for the focal length of the eyepiece lens, what weβll call π sub π.
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To begin our solution, letβs recall that the magnification π of a telescope system is equal to the focal length of the objective lens divided by the focal length of the eyepiece lens.
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Applying this relationship to our scenario, since the focal length of both our objective and eyepiece lenses must be positive, we can write that the magnitude of π is equal to π sub π divided by π sub π.
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Rearranging this equation to solve for π sub π, we see that itβs equal to the objective lens focal length divided by the magnitude of π.
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When we plug in for these two values and calculate the fraction, we find that itβs equal to positive 10.0 centimeters.
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This is the focal length of the eyepiece lens of the system.