a-grade-school-student-plans-to-build-a-35-power-telescope-as-a-science-fair-project-she-starts-with

Question

A grade school student plans to build a 35 -power telescope as a
science fair project. She starts with a magnifying glass with a focal
length of 7.5 $\mathrm{cm}$ as the eyepiece. What focal length is needed for
her objective lens?

Nathan Silvano · Accepted Answer

So in this problem, we have until the school with a total magnification. That&#x27;s called M of 35 times, and this telescope uses on eyepiece lens with a focal length. Let&#x27;s call the focal eyepiece off 7.5. Same team enters, and we want to discover difficult, objective lens. Well, we know that&#x27;s by definition, magnification is just the focal objective divided by the focal eyepiece. Therefore, we can we can just corporal weekend describe that the focal objective is just the magnification times, bifocal eyepiece. So this is going to be equal to Let&#x27;s see the focal objective in Kloster dirty five times judo 0.7 five. So this is just to a point six meters, and that&#x27;s the answer to this problem. Thanks for watching.

Netra Sharma · Answer

So in this problem, the length of the telescope is 47.5 centimeter. The focal length of the I piss is 1.5 centimeter. So in part A, we&#x27;re going to find the focal length of the objective. And to do that, we&#x27;re going to use that the length of the telescope is actually the sum of the focal length of the objective and focal length of the pipes. So from here, F or is going to be l minus F E, which is going to be 47.5, minus 1.5. It calls 46 inhibitor now in part B. We want to find out the angular magnification of the telescope. To do that. We just used the definition of the angler magnification, which is F over f e. And we have the value for the boat, which is 46 over 1.5, and this would give me 30 0.67 And that&#x27;s gonna be the magnification here. I was just ride down 30 approximately

Nathan Silvano · Answer

So in this problem, we have a telescope with a magnification off 75 times. We have a focal eyepiece land Focal. All right, peace lands off. Let&#x27;s see, I&#x27;ve seen teenagers five centimeters and we want to discover the Barrow Lentz off this telescope. Well, we know by definition, that barrel ins is just this some off the objective focal with the eyepiece focal. But the problem here is that we do not have the objective focal yet. So we need to use the magnification equation, which is F objective, divided by f. All right, he&#x27;s to discover the F objective, and then we can calculate the barrel length. So a skull Cletus the focal objective. It&#x27;s just going to be 75 times 0.5 which is the eyepiece focal in leaders. Okay, so this is just three point seven five meters. Now that we have the objective focal, we can calculate a barrel length. It&#x27;s just going to be three point 75 plus 0.5 So this is equals two three point eight meters. And that&#x27;s the barrel and off the tell the school. Thanks. For what

Mayukh Banik · Answer

the power or magnification power off a compound microscope is given by m equals. We&#x27;re going to use the absolute value. Equus as over focus went up the objective thence 25 centimeter over if off the eyepiece. Okay, in this particle, that problem I am supposed to find the focal point of the eyepiece. So focus on that. The IBIs becomes 25 centimeter times the land pit off the objected toe eyepiece distance which is given to be tense. And Demeter Eddie Quist, 10 centimeter, divided by objective focal event, which is given by 4.5 millimeter. Yeah, times 300 is the power. So that comes out to be 25 centimeter time stand centimeter divided by 0.45 centimeter times 30. And if I do the math, it comes up to be 25 over three times 0.45 centimeter, 18.5 centimeter. So focused and at the eyepiece has to be 18.5 cent

Carlos Henrique De Lima · Answer

we have astronomical telescope that has objective with focal 0.75 centimeters and plus 35 the eyepiece. And we want to know what is the total magnification. So the magnification can be reading us miners F zero and the pull over off the less and there&#x27;s this minus 0.75 meters times the power 35 b and this gives us a magnification off minus tiny six.

Surjit Tewari · Answer

the magnification is given this 3.5 and we know that magnification is equal to the year point upon the focal length. So and by physical too three point by there. And is the near point off 25? Sending me done they were aided by is equal to 3.5 solving this. We will get the for Berlin as it will do seven point 14 centimeter.

Problem

A 75 -power refracting telescope has an eyepiece …

Problem 69 Easy Difficulty

A grade school student plans to build a 35 -power telescope as a
science fair project. She starts with a magnifying glass with a focal
length of 7.5 $\mathrm{cm}$ as the eyepiece. What focal length is needed for
her objective lens?

Answer

$=3.6 \mathrm{~m}$

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$=3.6 \mathrm{~m}$

Physics

Marshall S.

Farnaz M.

Zachary M.

Jared E.

Wave Optics - Intro

Multiple Aperture Interference - Overview

James S. WalkerPhysics

Marshall S.

Farnaz M.

Zachary M.

Jared E.

James S. Walker

Physics