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Astronomers often take photographs with the objective lens or mirror of a telescope alone, without an eyepiece. (a) Show that the image size $h^{\prime}$ for a telescope used in this manner is given by $h^{\prime}=f h /(f-p),$ where $h$ is the object size, $f$ is the objective focal length, and $p$ is the object distance. (b) Simplify the expression in part (a) if the object distance is much greater than the objective focal length. (c) The "wing-span" of the International Space Station is $108.6 \mathrm{m},$ the overall width of its solar panel configuration. When it is orbiting at an altitude of $407 \mathrm{km},$ find the width of the image formed by a telescope objective of focal length 4.00 $\mathrm{m} .$

a. h^{\prime}=\frac{f h}{f-p}

b. h^{\prime}=-\frac{f h}{p}

c. 0.00107 m

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in this problem to solve Part A, which is to actually find the height of the image formed by the objective of the telescope. We actually used the lens equation one of her half, because one of our P plus one of our cue and from here Q is going to be cool, too. PF over p minus f Let's call this equation one now, using the definition off lateral magnification of the objective M, which is the lateral magnification of the objective of the telescope, it is going to be each prime over H or negative to overpay. So now, using the value of cue from Equation one, we can actually write down his primary H equals minus F over P minus F. And from here, we can actually see that h prime is going to be called to minus F H over T minus f. If your rear rinse, then it's gonna be a H over F minus peas. Okay, so this is part of a no to go to part two. We're just going to use the limiting case here, R B. So we're using the telescope. So in this case, he is going to be very very greater than for collect up the up sexy. So that means half minus p is actually going to be cool. Thio Negative p So now, from equation to here, we can actually write down each prime equals negative if age over P. So that's the solution to part B. Now, for Percy, the size of the object is given, which is hh oh equals 108.6 meter. An object distance is also given, which is P cause 407 kilometer. Let me write down in terms of meter and objective focal length is for meter. Now since the object is the distant object computer, the sides of the telescope we used their relation from part B and ignore the focal length in the den and in the numerator. So this is gonna give us a TSH prime because negative f o a spo overpay. And this is gonna be minus four times 108.6 over 407000 And this is going to give us negative 1.107 negative 0.107 times 10 to the negative three meter so you can see the negative valley here that end against the image is inverted

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