Two converging lenses, one with ƒ = 4.0 cm and the other with ...

Question

Two converging lenses, one with ƒ = 4.0 cm and the other with ƒ = 41 cm , are made into a telescope. (a) Determine the telescope’s length and its magnification. Which lens should be the eyepiece? (b) Assume these lenses are now combined to make a microscope; if the magnification needs to be 125 x , how long would the microscope be? Explain why the result in (b) is so impractical.

Accepted Answer

Welcome back. Everyone in this problem, Emma has two magnifying lenses, one with a focal length of five centimeters and the other with 35 centimeters. And she decides to assemble a simple optical device, identify which lens should act as the IPs, calculate the magnification and determine the telescope's length. Calculate the length of the microscope and explain why this set up would be impractical if she used these lenses to create a microscope with 150 times magnification. A says the lens with a smaller focal length is the eye piece. The telescope's magnification is negative seven times the telescope length is 40 centimeters. The microscope length is 1.05 m and it's impractical because microscopes are typically around 20 to 30 centimeters. B says the lens with the smaller focal length is the IP the magnification is negative six times the length is 40 centimeters. The microscope's length is 11 m and that's impractical for the same reason, microscopes are typically around 20 to 30 centimeters. C says the lens with the larger focal length is the IPs. The magnification is negative seven times and the length is 50 centimeters. The microscope's length is 10 m, which is impractical because microscopes are typically around 20 to 30 centimeters. And this is the lens with the smaller focal length is the I the magnification is negative seven times the length is 40 centimeters. The microscope length, the microscope's length is 11 m, which is impractical because microscopes are typically around 20 to 30 centimeters. Now let's try to start by figuring out the magnification and when we calculate the magnification, that will help us to figure out which I sorry, which lens should act as the eye piece. OK. Now recall that the magnification M is calculated as the negative value of the objective focal length divided by the focal length of the IPs. OK. Now, if we think about it, OK, if we think about it, if we are going to get a magnification or the magnitude of the magnification is going to be greater than one. OK. If the magnitude of the magnification is gonna be greater than one, what does that tell us? Well, it tells us then that the focal length of the objective lens, OK, would have to be greater than the focal length of the IP lens. So that means it would make sense for us to use the lens with a smaller focal length to achieve a magnification greater than one. OK. So let's make a note of that the lens with a smaller focal length, OK, which is five centimeters is chosen as the eyepiece. OK. Now that we know that we can go ahead and find the magnification because no, this tells us then that magnification is going to be the negative value of 35 centimeters divided by five centimeters which equals negative seven. Therefore, the telescope's magnification is negative seven times where negative tells us that it's an inverted image. No, how can we find a telescope length? Well, the telescope length, telescope length, which we can call L OK. Telescope length L is going to be equal to the focal length of the objective lens plus the focal length of the IPs lens that's 35 centimeters plus five centimeters which equals 40 centimeters. So we've solved the first part of our problem. OK. Now, we need to figure out the microscope length and from it, we need to explain why this setup would be impractical to create a microscope with 150 times magnification. Now, what do we know about the length of a microscope? Well, recall that the formula for the length of the microscope is equal to, tells us that the magnification equals negative NL OK. Divided by fe multiplied by fo OK. Thus, if we're going to solve for the length L OK, it's gonna be equal to negative MFEFO divided by N OK. We're here, we can try to figure out our value. OK. Now, in this case, if we substitute our values, OK, that means L the length of the microscope is going to be equal to negative 150. That's the magnification we chose multiplied by five centimeters, multiplied by 35 centimeters divided by 25 centimeters. And when we were, at that thought, it's equal to negative 1050 centimeters or its absolute value is approximately 11 m to two significant figures. Now, that tells us that the length of the microscope would be 11 m, but that's impossible. That's impractical because microscopes are usually around 20 to 30 centimeters. Thus, to sum up all we've done. Ok. Then the lens with the smaller focal length is the IP the telescope's magnification is negative seven times its length is 40 centimeters. The microscope's length is 11 m. And we know that that's impractical. If we look back at our answer choices, then that tells us d is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Lens Formula

Magnification

Telescope and Microscope Design

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