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Predict/Calculate You are taking a photograph of a horserace. A shutter speed of 125 at $f / 5.6$ produces a properly exposedimage, but the running horses give a blurred image. Your camerahas $f$ -stops of $2,2.8,4,5.6,8,11,$ and $16 .$ (a) To use the shortest possible exposure time (i.e., highest shutter speed), which $f$ -stopshould you use? (b) What is the shortest exposure time you canuse and still get a properly exposed image?

a) 2b) $1020 \times 10^{-6} \mathrm{m}$

Physics 103

Chapter 27

Optical lnstruments

Wave Optics

University of Washington

Hope College

McMaster University

Lectures

02:51

In physics, wave optics is…

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Interference is a phenomen…

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You are taking pictures of…

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Okay, So in this problem you are taking a photograph of a horse race and you have a shorter speed enough number. But this configuration gives you, gives you a blurred image. So when the first item of the problem, we need to explain what should be the best f stop. It gives us the best image in this scenario. Okay, First of all, we know that Let's put in here. We know that your teacher diameter is just the focal aunt divided by the F number. So we can see in this formula that the smallest F number gives us the largest are featured a emitter. So if we have a bigger diameter, that means in the lens of the camera that we're going to have the bigger area. So if we have the bigger area, we can capture more light. And with more light, we have a better image. We do not have a blurred image. So following this thinking, we can say that in the first item of the problem, the best F stop that gives us the best image is there Stop too. All right. In the second item of the problem, it's pretty in here the second night end of the problem, we need to calculate what is the shortest exposure time we can use and still get a properly exposed image. Okay, so as we know, delights are captured by the area off the lens. So if we make on area final area off the lens divided by the initial area off the lens and we fit calculated this we need to know. We can discover how much the area needs to change to get the better image. This is going to be necessary to understand how much exposure time when we're gonna need to solve this problem without changing the our feature diameter. Okay, so the area of the lenses a circle. So I was just going to be one divided by four time spy times. The our Teacher Diameter square. This same for the initial one. I'm spy times in the show. Our preacher diameter square. We can cancel. Eat this too. So the area is just the final diameter square divided by Deanie Show Diameter square. That's calculated the other page. Okay, we know that we know from here that the diameter is just a focal then divided by the F number, the focal in off the camera doesn't change. So that's put in here. This ratio is just going to be one divided by two square because this is the final stop that we calculate. And the initial one is one divided by 5.6. It's square. So this is after calculating it's going to be equal to seven point eight. So this is there, Rachel off the change of the area so we can have the but the best image possible in this situation. So if if we do not want to change the diameter, we need to change the exposure time, and we know that our exposure time is just one divided by 1 25 seconds. So this is the initials Bowler Time. So our new exposure time it's called a final exposure time needs to decrease by seven points. Eight. So is just one divided by 1 35 times one divided by 7.8. This is equals to one divided by nine Sorry, 975 which gives us approximately one divided by 1000. So this is the explorers

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