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Problem 66 Easy Difficulty

A laboratory (astronomical) telescope is used to view a scale that is 300 $\mathrm{cm}$ from the objective, which has a focal length of $20.0 \mathrm{cm} ;$ the eyepiece has a focal length of 2.00 $\mathrm{cm} .$ Calculate the angular magnification when the telescope is adjusted for minimum eyestrain. Note: The object is not at infinity, so the simple expression $m=f_{o} / f_{e}$ is not sufficiently
accurate for this problem. Also, assume small angles, so that tan $\theta \approx \theta$

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Top Physics 103 Educators
Christina K.

Rutgers, The State University of New Jersey

Andy C.

University of Michigan - Ann Arbor

Marshall S.

University of Washington

Zachary M.

Hope College

Video Transcript

{'transcript': "so we need to solve this problem in a numerous part. So we start by first analyzing the image formed by objective. So we have the distance off. The object from the lens is 114 meters. We know that the focal length is 1.5 centimeters and we will try to find the position off the image. So using the thing lands Formula One over the old plus one over t. I is equal to one over f. We have one over 114 plus wine over T. I is equal to 1/1 0.5 or one or G. I is equal to 1/1 10.5, minus one over 114. So that gives us D. I is equal to 1.52 meters. So we cannot find the magnification off the objective, which is negative. G I over deal so negative 1.52 divided by 114 used us to magnification off the objective as negative 0.133 So now that we know the magnification off the objective bugle call head and find the discuss, the image formed by the eyepiece. Okay, so we know that deal. The distance off the architect will be equal to l minus 1.5 to where l is the distance between the two lenses and that is actually equal to the sum off the two focal. And so this will be F all plus F e minus 1.52 The F four is 1.5. The F e is zero point your seven minus 1.52 So the Arctic distance for the objective comes out to be 0.5 meters. And you already said that the focal length is zero points Europe seventh beakers. So we start by finding out the location off the image. So one over the O plus one over the eyes, equal to one over F 1/0 0.5 plus one over the I is equal to 1/0 point. You're seven or one over. The I is equal to 1/0 points. Your seven minus one overview apart your fight. So that gives us d i his equal two negatives. Your point wanted eight meters so we can now find the magnification off the eyepiece, which is negative. D I over deal so negative off negative 0.18 divided by 0.5 So that comes out to be negative. That comes out to be positive. 3.6. So now we know the ratification of the objective. We also know the magnification off the eyepiece. So the overall magnification and would be equal to have, oh, times and e. So we had em or is equal to negative for 0133 and then times that with 3.6. So we get the overall magnification is equal to negative 0.48 So this is the linear magnification. Now let's go ahead and find the angular magnification. So far thing for angular magnification that we need is the angle form by the object on the unaided eye. So that would be equal to HBO. Over D O and T O is equal to sew age, for we don't know, but D O is equal to 114 plus zero points your seven plus 1.5. So that's the distance off the object from the my telescope, plus the some off the two focal ends so that's equal to H over. So we can add those numbers together. So 114 plus points or seven plus 1 25 comes out to people 215.57 Next, we need up the image, the angle of an image form, windy telescope. So that's going to be able to h I over D I and D. I just figured out was equal to zero point. I want it. So each eye over D I and G I we just figured was your point 18 meters. So the angular magnification m angular is equal to fatal crime over Sita. Okay, so this is equal to negative. Oh, since study images inverted so this would be negative. So that would be negative. H i over 0.1 it divided by H four over 115.57 So that comes out to be equal to negative each eye over each all times 115.57 divided by 0.18 But h high over h O is actually equal to the overall magnification, which we found was your 00.8 So this is negative 0.0 for eight times 115.5 7/0 0.18 and that is equal to negative 31."}

Other Schools
Top Physics 103 Educators
Christina K.

Rutgers, The State University of New Jersey

Andy C.

University of Michigan - Ann Arbor

Marshall S.

University of Washington

Zachary M.

Hope College