The focal length of the eyepiece of a certain microscope is 18.0 ...

Question

The focal length of the eyepiece of a certain microscope is 18.0 mm. The focal length of the objective is 8.00 mm. The distance between objective and eyepiece is 19.7 cm. The final  formed by the eyepiece is at infinity. Treat all lenses as thin. (b) What is the magnitude of the linear magnification produced by the objective?

Accepted Answer

Hi everyone. In this practice problem, we're being asked to determine the strength of a linear magnification produced by the first lens of a microscope. In a science project, a student built a microscope with two converging lenses. The first one formed the microscope's objective with a focal length of 1.5 centimeter. The second one forms the microscope ocular with a focal length of 4.5 centimeter. The first lens forms a real image exactly at the focal length of the ocular. And the two lenses are 8.5 centimeter apart, we are being asked to determine the strength of the linear magnification produced by the first lens. And the options given are a 0.5 B 0.6 C 1.67 D 2.0. So in this practice problem, the image formed by the first lens is going to be located at the focal length of the ocular or A F two. Therefore, because of this, the final image will be located at infinity. So the image distance S one prime will be equals to or formed by the first lens is going to be placed four centimeter to the left of the first lens. So in this case, s one prime is going to be equals to the distance of the two lenses which is 8.5 centimeter minus the focal length distance or the focal length length of the first of the second lens, which is going to be 4.5 centimeter. In this case, then S one prime will then be equals to 4.0 centimeter to make this an easier to make this an easier representation. I'm going to draw our system or the system of the two lenses of the microscope, we'll have L one here and we'll have L two located here. I'm gonna uh label that L one and L two where L one, we have F one and F one prime here. And then for L two, we have F two and F two prime. The distance between L one and L two is 8.5 centimeter. And the distance between L one and F two or S one prime is going to be four centimeter because in this case, the focal length or F or F two, the distance from the focal length to L two is 4. centimeter. So this is 4.0 centimeter just like that. Awesome. So now what we can do is we can actually move on and calculate S one or the object distance found using the object image relationship. In this case, we have S uh one divided by S one plus one, divided by S one prime equals to one divided by F one. So this is the object image relationship. And it is given in the problem statement that our F one or F one prime distance is 1.5 centimeter. S one prime is given to be 4.0 centimeters. So through that we can rearrange our equation to get an equation for S one. So the S one wheel equals to F one multiplied by S one prime divided by S one prime minus F one. So we can substitute all of our values into this equation right here to get the object distance as one. So this will then be 1.5 centimeter for F one multiplied by 4.5 oh 4.0 centimeter for S one prime divided by 4.0 centimeter for S one prime minus 1.5 centimeter for F one. This will give us our S one or object distance of 2. centimeter. Next. Now that we have obtained our object distance, we can then actually calculate the linear magnification produced by the first lens because the linear magnification or M one will follow the formula of negative S one prime divided by S one, we can substitute all this value. So S one prime earlier is 4.0 centimeter. So then our M one will then equals to negative 4. centimeter divided by S by S one which is going to be 2.4 centimeter. And then calculating this, we will then get our uh linear magnification of the first lens to then be equals to negative 1.67. And then the strength of the linear magnification will then just be equals to the absolute value of M one or 1.67. So there we have it, the strength of the linear magnification will be 1.67 which will correspond to option C in our answer choices. So that will be it for this video. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on our website and that will be it for this video. Thank you.

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