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University of Wisconsin - Milwaukee

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Problem 5 Easy Difficulty

A camera is being used with a correct exposure at $f / 4$ and a shutter speed of $\frac{1}{15}$ s. In addition to the $f$ -numbers listed in Section $25.1,$ this camera has $f$ -numbers $f / 1, f / 1.4,$ and $f / 2$ . To photograph a rapidly moving subject, the shutter speed is changed to 1$/ 125$ s. Find the new $f$ -number setting needed on this camera to maintain satisfactory exposure.

Answer

1.4

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Top Physics 103 Educators
Elyse G.

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Farnaz M.

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Video Transcript

So in this problem, we actually have a camera initially set of for F number calls for and the exposure time of the camera is 1/15 seconds. Now we want to take a picture of a fast moving object and then for that case, we actually wanna decreased exposure time. So what we do is we change it to one over 125 seconds. So now for that case, we want to find out the F number, the final F number, right? So let's begin. The first thing that we need to know here is initially it was set up at correct exposure. And we want to change the F number in such a way that we are still getting the correct exposure. So what does that mean physically? So what that means is, in both of the cases initial and final cases, the energy that is being delivered to the film of the camera is the same. All right, sufficient energy so that you get the correct exposure. So let's begin with that. So the energy let's say e one is the initial energy has to be cool to eat too. Both of the energies that are being delivered to the film. All right, so now, from the definition of the intensity intensity, I is actually the energy per unit area per unit time. All right. So from here, we can actually re errands and right for the equation here for this e one equals two. We can write this down. This is gonna be I won times a times t one because I to time's a times T too. So do you see a here? I didn't write it a one and a two. It is because the area of the film is the same. So from here, we can actually right down in next page. I won t one equals I too. Tea too. Great. So it is I want you want because I too tee too. Now I'm gonna rewrite this. I want over I too, right. I'm trying to write. I won over I too. I to because t one over tea too. Never main. This is gonna be t to over t one two to over t one. So now we want to remember the relation between the intensity and deaf number. So if you remember, intensity is actually inversely proportional to the square of the F number. Right? So if you want to write down, I want and I too. In terms of that, we're actually going to get this. Have to f number two at number two, squared over half. Number of one squared. This is squared. So So you know, after over two square over F number one square is going to be cool, too. T to over t one. And we know the value of T one and tea, too. T one was won over 15. And t two is 1/1 25. So if you calculate that you're going to get 3/25 All right. So what we want to do from here is from here, You can see, actually F number two is going to be the square root off 3/25 times. Half number one. All right, so this is going to be cool, too. 1.4. So this is the final laugh number

University of Wisconsin - Milwaukee
Top Physics 103 Educators
Elyse G.

Cornell University

Farnaz M.

Other Schools

Zachary M.

Hope College

Aspen F.

University of Sheffield