the-actual-light-sensor-size-of-a-digital-camera-is-149-mathrmmm-times-223-mathrmmm-you-want-to-take

Question

The actual light sensor size of a digital camera is 14.9 $\mathrm{mm} 	imes$
22.3 $\mathrm{mm} .$ You want to take a photo of your friend, who is 1.9 $\mathrm{m}$
tall. Your camera has a $55-\mathrm{mm}$ -focal-length lens. How far from the camera should your friend stand in order to produce a $22-\mathrm{mm}$ -tall
image on the light sensor?

Nathan Silvano · Accepted Answer

Okay, so in this problem, your friend, uh, taking a picture, Let&#x27;s see if they can&#x27;t picture and he want we want to discover a reach distance off the camera. That means that these you know, he needs to stand so he can produce image off 22. You&#x27;ll meet her stall. We know that the height off our friends is swamp 0.9 meters. And we know that the focal lens of the camera, it&#x27;s just 55 of the meters. Okay, so how are we going to calculate this? We know from chapter 26 there is a way to combine distance and height using the magnification. So we know that distance of the image is just going to be miners magnification times zero. And we know the magnification is just height of the image, divided by height off the object. So combining this too, we have the distance off. The image equals to minors. H I divided by 80 times dessert. Okay, well, this is not enough to solve the problem because we do not have this sense of the image so we can use Phelan&#x27;s a creation that is one divided by F. If course one divided by I lost. Oops, Sorry. One divided by zero. One divided by the zero. And we can substitute the d I that we have here. So this is going to be quote too. Let&#x27;s see. One divided by F E clothes a zero divided by minus age. I do you zero well US one divided by zero. If we put these one divided by the zero evidence, we&#x27;re going to have one divided by zero the multiplies one minus 80 divided by h i. So to blind is in cross. We have that, do you? Zero It&#x27;s just going to be f multiplies one miners, 80 divided by age one. Now we can finally calculate inserted the distance. Okay, so substituting Oh, dividers we have here we have that The zero is 0.0 55 because we need to convert millimeters two meters. So this is going to be 0.0 55 times one miners, 1.9 divided by minus. Divided by minus 0.0 22. So this is just the zero equals to 4.8 meters. We put a minor sign in age I in here. If what this minor sign in here? Because the image is inverted. And when the image is inverted, we need to put a negative sign. The calculation. Therefore, the minus sign is there. Okay, so this is the final answer.

Bruce Edelman · Answer

Okay, so we&#x27;re doing Chapter 34. Problem 47 here. So it says when a camera is focused, the lens away or toward the digital digital sensor. And if we take a picture of our friend who is ending 3.9 meters from the lids, so that means our object sincerely. 3.9 meters using the camera with they limbs that 85 millimeter focal. Okay, um, how far away from the sensor is balloons? Cool. So we know that we know S and F so we could just obviously solve for that. Sprint was Put this in meters first. Such 0.0 a five years. Be arranging the thin Lindsay equation. See, that s S f over us minus. This is 3.9 times 0.85 over 3.9 minus reports your five. Cool. So let&#x27;s plug this in and Saul, and we should see, comes out too. Zero port 0869 meters. So that means that the senses is 86.9 millimeters from villains. Cool. Okay, so the next part it says, um will the whole image of your friend who was 175 centimeters. Call fit on the sensor. That is 24 by 36 millimeters. Okay, So that means if our object height is 175 centimeters tall, we want to figure out if the image will fit on that screen. Right? This is a screening size. Okay. So again from the magnification, we know that the negative image distance over object distance equals immense high over object height rearranging. This is negative. Best prime over. Yes. Times. Why? Cool. So we just need to plug in here. So negative. 0.869 3.9 times 1.7. Plug. Aladdin. We should see this comes introduced you to to do that being negative. 39 millimeters. Cool. So are images inverted, but it&#x27;s 39 millimeters tall with the screen high only being 36 meters. So no, it will not sit. No, it will not. Okay, Cool

Dading Chen · Answer

also for the first question when you&#x27;re determining the object is since which is the whole Fars for your friends stand from Okay, so we know. Ah, well over four glances people want over image distance my as well. Over I&#x27;m distance and the former for magnification is equal to minus We over you just also able to hpai over age It&#x27;s refines the height of image edges the actual hearts so far in question were not me. Hi. The image is 36 millimeter and the actualize 2.0 meter. So if we do, some arrangements will get Actually, here is a list of this be is even toe minus H five over age task You okay? And my s h pry over age can be simply legal too. Minus 36 times tend to about off next three. Peter, you seem salon millimeters Tend to apartment three meter Okay, over H, which is 2.0 meter test. You okay? This will give us He is equal to minus 0.1 a You okay? And if we plugging back to the equation that we that I give you guys a first, we&#x27;ll get one over f was just one over. The focal lens is simply equal to one over zero. Ah, minus zero point zero on a you minus one over you. Okay. Used print. This is here, and this will give us our f is equal to one over you. Times one over. Minus 0.0 on a minus one. Okay. And since we know they vocal ins, it&#x27;s Caymus. 50.0 millimeter. Which chemicals? Or 50 times? 10 to the power of 93 meter. So you can be equal to ah one over minus 0.1 a minus. One times the focal A. Okay. And this will give us one over minus 0.0, while a minus one times 50. Ties tend to Ah, our language. Three meter. You legal too? Which is object distance only. Ego as minus two point. Any three meter. Okay, we&#x27;ll question be it was asking for the image distance in a question. Which is how far is the land spawned appeal. Okay, so we gotta figure out be and remember from a previous question, me is equal to minus 0.1 a. You and we&#x27;re in kind of you you is given as minus to appoint a three meter, which means that the image this is we can be you go to minus 0.1 a times minus 2.83 meter, which is you go to 0.0 if I won meter. If we convert it to Sam, meter is 5.1 centimeter. Okay? And that&#x27;s the solution for this question. Thank you.

Sheh Lit Chang · Answer

these questions you have given the situation whereby when that camera its focus Ah, the lenses move away from all toward the film king. So then we are using this camera to take, uh, picture of your friend sending 3.9 meters on the lens using lands camera lens off 85. Mm. Focal ing. And we want to find ah, image distance. Okay. And then whether the image if it&#x27;s no film so how it would be using first to find an image distance, we&#x27;ll be using one of s plus on the S. Crime goes to one over F. So when the U. S prime this one over, ask minus one over R s. Hey, uh, is 85 uh, times 10 to the minus tree, Meet us 85. Mm. And then s is, uh, 3.9 meters. So why no s prime iss 1/85 times. 10 to the minus. Tree minus 1/3 100.9. Uh, so you get asked prime to be, uh, you could too. Your 0.869 meters. The heights. Oh, unit. Yes. Uh huh. Magnification times, huh? Object height. So B s prime over s times. Why? 0.869 right. By, uh, three point ny times. Why, if 175 cm Hey, and then solving this, you get three point night. See him? So this is, uh, greed. Er then 36 mn. So the image that&#x27;s no feet. Eso Okay, so this is the, uh, um distance from soon to cleanse and then that you each that&#x27;s not feet that film.

Farhanul Hasan · Answer

all right. We want an inverted image, which means that magnification will be negative. Ah, and so we can work out magnification and is image height over object height and and so image height will be negative. 0.8482 um, centimeters, because that&#x27;s 4.82 millimeters. And you divide that by 4 50 centimetres to get magnification of negative one point of seven times 10 to the negative three look, okay. And so we know that M is also equal to negative image. Distance over object distance. So? So this means that image distance as prime as one point. There are seven times 10 of negative, three times s. Okay? And then we use them. The familiar focal length equation, one of her focal and people&#x27;s one over image. This object distance was one of her image distance, So Ah, so this well, so this will be equal to one over us. Plus one over, um, this one over 1.77 times 10 to the negative Me times s onda. And so you can take You can take as common here, worked that out and take in verse to get well. What&#x27;s not F, but s is what I meant to write a Z call 29 34 points. Six times Focal int which were given is 50 millimeters or five centimeters. So this gives us 4,670 centimeters, which is equal to 46.7 centimeters. So you have to place your film. You have to be 46.7 centimeters aware from the giraffe.

Bruce Edelman · Answer

Okay, so we&#x27;re doing Chapter 34. Problem 45. So this from says a camera lens has the focal links of 200 millimeters, and it says how far should bear how far from the limbs should the subject for the photo be if the lenses 20 centimeters from listens. So if the lenses 20 centimeters from the sensor, that&#x27;s where we want the image to be formed is on the sensor. So our s prime, the image distance will be 20.4 centimeters. Cool. So if we know this in the local links, we can obviously just use thesis and Lindsay equation. Awesome. So what we&#x27;re, um, trying to figure out is obviously that&#x27;s the one we don&#x27;t know. We rearrange. We see that this is us prime f over s firing minus F so we can easily plug that in. So this becomes a 20.4 times 200. Let&#x27;s, uh, probably need to use these in the similar units. So this is 200 millimeters, which is the same as 20 centimeters. So here&#x27;s this guy. Cool. So 20.4 times 20 over 20 points for minus. So what is this equal? We got four times misuse. Cool. This comes out to being 10 27 years. So this is 10.2 meters. That&#x27;s a very far away object. So yeah, so But the key point to this is in this camera. If we have a lens with the focal length of around 20 centimeters, that&#x27;s correspondent Thio, taking good pictures of subjects 10 meters away. Cool.

Problem

(a) Find the $f$ -number of a telescope with an o…

Problem 10 Easy Difficulty

Answer

4.80 $\mathrm{m}$

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