### Video Transcript

A 7.5x binocular produces an angular magnification of negative 7.50, acting like a telescope. Mirrors are used to make the image upright. If the binoculars have objective lenses with a 75.0-centimeter focal length, what is the focal length of the eyepiece lenses?

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. 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. We want to solve for the focal length of the eyepiece lens, what weβll call π sub π.

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. 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.

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. When we plug in for these two values and calculate the fraction, we find that itβs equal to positive 10.0 centimeters. This is the focal length of the eyepiece lens of the system.