A 0.150-centimeter focal length microscope objective is 0.155 centimeters from the object being viewed. What magnification is produced? What is the overall magnification if an 8x eyepiece, one that produces an angular magnification of 8.00, is used?
Lets start by recording some of the information we’ve been given. We’re told that the objective lens of a microscope has a focal length of 0.150 centimeters; we’ll call that 𝑓 sub 𝑜. We’re also told that the object that the microscope is viewing is 0.155 centimeters from the objective lens; we’ll call that 𝑑 sub 𝑜.
We want to know the magnification produced by the objective lens, which we’ll call 𝑀 sub 𝑜. We also want to know the overall magnification if an 8x eyepiece is used; we’ll call that overall magnification 𝑀 sub 𝑠, the magnification of the system. Now at the start, our system consists of two objects: the object itself and the objective lens.
The lens creates an image, and we want to know what is the magnification of that objective lens. Let’s recall some magnification rules. The magnification of an optical component, 𝑀, is equal to the height of the image it produces divided by the height of the object producing that image. And it also equals negative the image distance divided by the object distance.
So in our case, the magnification of the objective lens is equal to negative the image distance created by the lens divided by the object distance. We don’t yet know the distance between the image created by the lens and the lens. But before we get to that, let’s recall one other fact about magnification. And that is that the total magnification of a system of optical components, where there are 𝑁 components, is equal to the product of the magnification of each component.
In other words, the total magnification is equal to the magnification of component one times the magnification of component two and so on and so forth, all the way to the last component in the system. This relationship will be useful to us in part two, but now we want to solve for the image distance of the image created by the objective lens.
To do that, we can use the mirror equation or also sometimes called the lens equation, which says that one over the lens’s focal length is equal to one over the object distance plus one over the image distance. In our case, one over 𝑓 sub zero, the focal length of the objective lens, is equal to one over 𝑑𝑜 plus one over 𝑑𝑖.
If we algebraically rearrange this equation to solve for 𝑑 sub 𝑖, we find that it equals the focal length times the object distance divided by the object distance minus the focal length. When we plug in for these two values, 𝑓 sub 𝑜 and 𝑑 sub 𝑜, we find when we calculate this fraction that 𝑑 sub 𝑖 equals 4.65 centimeters.
So now we’re ready to plug in 𝑑 sub 𝑖 and 𝑑 sub 𝑜 into our equation to solve for 𝑀 sub 𝑜, the magnification of the objective lens. When we calculate this fraction, we find that the magnification is negative 30.0 times. In other words, the image created by the objective lens is 30 times
smaller [larger] than the object, and it’s inverted. Now that we’ve solved for the magnification of the objective lens, we imagine that we add another lens to our optical system, called the eyepiece.
And we’re told that the magnification of this lens, which we’ll call 𝑀 sub 𝑒, equals 8.00. To solve for 𝑀 sub 𝑠, the magnification of this overall system of two lenses, we’ll multiply the magnifications of the objective lens with that of the eyepiece. When we plug in for these two values, negative 30.0 for 𝑀 sub 𝑜 and 8.00 for 𝑀 sub 𝑒, we find that the overall magnification of this combination of lenses is negative 240 times. So the image that’s created is 240 times bigger than the object and it’s inverted.