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Camera with 443 mm focal length lens used to photograph the sun and moon What is the height (in mm) of the image of the sun on the film, given the sun is 1.40 km in...

Question

Camera with 443 mm focal length lens used to photograph the sun and moon What is the height (in mm) of the image of the sun on the film, given the sun is 1.40 km in diameter and is 1.50 km away? (Include the sign of the value in your answer.)4.13mm

camera with 443 mm focal length lens used to photograph the sun and moon What is the height (in mm) of the image of the sun on the film, given the sun is 1.40 km in diameter and is 1.50 km away? (Include the sign of the value in your answer.) 4.13 mm



Answers

A camera with a 100 $\mathrm{mm}$ focal length lens is used to photograph the sun and moon. What is the height of the image of the sun on the film, given the sun is $1.40 \times 10^{6} \mathrm{km}$
in diameter and is $1.50 \times 10^{8} \mathrm{km}$ away?

So the strategy of the question is you find G I It was one of the F minus one of our dio Put the values in and find d I now magnification is h I over h all. It was negative d I over Theo, so find the image distance to be negative. De i o r deal times h o. So that's all this We have the focal than this 200 millimeter are 0.2 m minus did not ease 10 kilometer So 10 to the 4 m which is almost 200 meeting Now we know that immense distances 200 m on the object distances can to the 4 m times height of the object He's 100 mean 1000 m So something this we get I'd up the image He's negatives to dance 10 to the negative two Major it was negative two centimeter So height of the image ease two centimeter on the negative sign in the guest that it is in the

A large is a major for Sung using different cameras for so for about a we're using, um 28 millimeters focal Leif. In general, all the cameras will have the same expression. So the because the sun is very far away, the land will be at the focal point. So the I is f and we can find the magnitude off the size. Using that age, I divided by 80 s minus t i by the zero. So the magnitude of the image is a zero. The eye over the zero. And you let supply this for this case. 28 millimeters. So we get a zero is 1.4 times 10 to the sixth communiqu times 28 millimeters, divided by 1.5 time. Stand to the eight planets. This is you know 0.26 millimeters for part eight for part, Be the same formal. But now we're using a lens that has 15 millimeters. So age one iss won't bring four times. Then to the 60. Liberace sounds 15 million mirrors divided by 1.5 times stand today, eh? Cleanliness. This is your 0.47 millimeters. This is the part Me and part. See, we were using 125 millimeters No mirrors. Now many mirrors so but see you're using a tch. One is 1.4 times into the six glimmers. I'm so happy that day. Five. Many mirrors divided by 1.5 times and to the eight. Then there's this 1.3 million years now for party. Ah, if the 50 million years ago is considered normal for this Lance, what what is the relative magnitude magnification Compared to the other two? Um, the equation serves that the image height is directly proportional to the focal life. Therefore, the relative magnification will be the racial off the focal, who, for the 28 million mirrors less because that is 28 by the by 15. This is there a 0.50 six. Therefore, the 125 million mirrors. This is 135 divided by 50. This is 2.7 times

So we're asked to figure out the size of an image on a camera for three different focal lengths of that camera, considering that the image we're looking at as the Sun, which has a diameter of 1.4 times 10 to 6 kilometers. So the original height of her object, H 70 is 1.4 times 10 to the six and it's had a distance. So the distance to the object is 1.5 times 2 10 of the eight kilometres. Okay, so because the sun is very far away, the image is going to be at the focal point of the camera. So we find the size of the image using Equation 23-9, we can solve for the height of the image from that equation. So for part A, we have M magnification is equal to the height of the image divided by the height of the object, which is equal to negative the distance to the image, which we said was just the focal length. Uh, well, actually, let's go ahead and write. Write it explicitly. So this is equal to negative the distance to the image divided by the distance to the object, which in our case is equal to negative the focal ing divided by the distance to the object. Therefore, solving for the height of the image, we find that this is equal to the, uh, negative height of the object times the focal length, which in the first case for part A is 35 millimeters. This is gonna be 35 millimeters divided by the distance to the object. So plugging those values into the expression, we find that for part a, the height of the image is equal to negative. So it's inverted 0.33 millimeters Okay, box, that is our solution for a part. B has asked us to do the exact same thing, but simply changed the focal length of the camera. So now, for part B, the hide of the image is going to be equal to negative height of the object which isn't going to change times 50 millimeters, in this case, divided by the distance to the object, which again doesn't change. So carrying out this expression, we find that this is equal to negative 0.47 millimeters, which we can therefore box in as the solution for Part B and now, lastly, part see again, we just simply change the focal length here. Everything else is the same, so the height of the image is going to be equal to negative height of the object multiplied. By this time, the Focal Inc there's 105 millimeters 105 millimeters divided by distance to the object. So carrying out this expression, we find that the height of the image in this case is negative. 0.98 millimeters three, right? That nine. That wasn't very nice looking 0.9 a millimeters weakened box said it is our solution for part C.

So for looking question twenty three point nine, we find the relation between height and ah images. Since so hide off image override off the object should be equal to negative high dose distance of the image over distance of the object. Now the way the sign works is if the image is inverted, ma'am, it is inverted. Then we take negatives. Negative, sane you take me to sign, um, and for camera, the images always inverted so we can take Dia's negative for our case. So the final question becomes each eye is equal to It's not times d I over the knot and we know all the values. So for our case, is not is ah one point four times ten to the power six kilometers than this image distance is twenty eight millimetres and the optic distances, one point five times tend to the bar. Eight senators Now the reason we took ah the eye as twenty eight is because Thea absence object is very far away. We know that the object, the image that forms behind the lens, is always at the focal point. So that's why since our focal length is given, we can take off for billing as our image distance. So and that's but it gives us so That's party, and that gives us our H igh as point two six millimetres three point two six moneylenders. Now we can do the same calculation for R, B and C, so leave up to you to do the calculation. But it's the same on living little changes. Thie on DH for part B for fifteen limiter. Sorry for fifty millimetre as R. D. I. The height off the object with the zero point four seven moneylenders. And for see when we have a focal length off one thirty five millimeters, the height of the object will be one point three fingers. Okay, now we have to find out the magnification, so we know that fifty millimeter is the standard. So that means for twenty eight millimeters focal length, the magnification is going to be zero point five six x, so that means it will be smaller than fifty millimeters. And for one thirty five millimeters, this will be two point seven X. So it's it's going to be two point seven times fifty millimeter, two point seven time off, fifty millimeter for polite Thank you


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