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(a) Find an equation for the length L of a refracting telescope in terms of the focal length of the objective $f_{o}$ and the magnification $m$ . (b) A knob adjusts the eyepiece forward and backward. Suppose the telescope is in focus with an eyepiece giving a magnification of $50.0 .$ By what distance must the eyepiece be adjusted when the eyepiece is replaced, with a resulting magnification of $1.00 \times 10^{2} ?$ Must the eyepiece be adjusted backward or forward? Assume the objective lens has a focal length of 2.00 $\mathrm{m} .$

a. f_{o}\left(1+\frac{1}{m}\right)

b.0.020 m

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in this problem for part A to find the question for the length off a refractive telescope in terms of focal length of the objective F not, and the magnification M. Let's recall that the length the total length of the telescope is actually f not plus Effie. Okay. And the magnification of the telescope is f not over Effie. So from here, we can actually use f e equals f not over m f over him. And this is going to give us l equals f o plus f over m because F o. M plus one by m no. For part B for part B, it is given that the eyepiece is adjustable. It is also sued that the telescope is initially in focus with the magnification initial magnification. M wa tickles 50 all right. And the focal length of the objective is f o f zero equals too bitter. Therefore, the initial length of telescope from the part A If we use part, eh is going to be hell one I'm gonna call the initial length of l one calls F O and one plus one over and one an M one plus one is going to be 51. So fo is too and it's going to be 51 over a 50 and this is going to be 2.4 meter. So this is initial right now. What we're gonna do is we're gonna move the telescope so that the magnification changes. Okay, so when the eyepiece is moved so that the new magnification is em too. So let's call this final Curious when the eyepiece has changed so that the new magnification is m two recalls 100 then the new length would be l too. Which is again use part, eh? It's going to be f o him, two plus one over M too. In this case, empty was 100. So two 100 Sorry, this is 100 or one so two times 401 over 100. And this is going to be 2.2 meter, 2.2 meter. So now the difference in length, which is Delta l. It calls the final length to minors. Initial length. Everyone, which is going to be 2.2 minus 2.4 It's going to be negative 0.2 meter or negatives. Negative. Two centimeter. Okay. So you can see that the length of the telescope is decreasing. That means the telescope is shortened or we can also say it is pushed in.

University of Wisconsin - Milwaukee