A certain type of film requires an exposure time of 0.010 s with an $f / 11$ lens setting. Another type of film requires twice the light energy to produce the same level of exposure. What $f$ -number does the second type of film need with the 0.010 -s exposure time?
f / 8.0
So in this problem we have a type of film which we're gonna call film one. So for correct exposure, the exposure time for that is t wanna go 0.1 2nd and F number goes 11 to get a correct exposure. Now we have another one film which we're gonna call film to, which requires twice the energy for the correct exposure. When we said t to go 0.1 2nd So what we want to do is we will find the F number for the film too. Okay, so let's start. Remember the definition of the intensity, the definition of the intensity is I because energy per unit area for unit time. So from here, you can write down E, sir equals intensity times area times t. So what does this mean for the film won and film to is he won it cause I want ducked a the t one. And he too, because I to dot a dot t one because t too close to one. So I'm just gonna replace with that. So if you look at these two relations here, you're going to see that given that e two equals twic one is gonna be I too you calls twice I want, right. So which also means this also means I to over I won Because to All right, so now, to find the f number two, we're gonna use the definition of the intensity in Trump's off half number. So remember, intensity I I is inversely proportional to the square of F number. So this means I too over I won. Actually is gonna be cool too aft number off. One squared over. Asked number off, two squared. And remember, I too over I want is equal to two. Right? So from here, you can see that half number of two is going to be equal to one by two, one by square root of two times after number off one, which is actually 7.8. So if we want to change the film to film too, then the actual F number is going to be cool too. Half over eight for the camera. Okay?