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Light enters the eye through the pupil and strikes the retina, where photoreceptor cells sense light and color. W. Stanley Stiles and B. H. Crawford studied the phenomenon in which measured brightness decreases as light enters farther from the center of the pupil. (see the figure.)

They detailed their findings of this phenomenon, known as the Stiles-Crawford effect of the first kind, in an important paper published in 1933. In particular, they observed that the amount of luminance sensed was not proportional to the area of the pupil as they expected. The percentage $ P $ of the total luminance entering a pupil of radius $ r mm $ that is sensed at the retina can be described by$$ P = \frac{1 - 10^{-pr^2}}{pr^2 \ln 10} $$where $ p $ is an experimentally determined constant, typically about $ 0.05 $.(a) What is the percentage of luminance sensed by a pupil of radius $ 3 mm $? Use $ p = 0.05 $.(b) Compute the percentage of luminance sensed by a pupil of radius $ 2 mm $. Does it make sense that it is larger than the answer to part $ (a) $?(c) Compute $ \displaystyle \lim_{r\to 0^+} P $. Is the result what you would expect? Is this physically possible?

Source: Adapted from W. Stiles and B. Crawford, "The Luminous Efficiency of Ray Entering the Eye Pupil at Different Points." Proceedings of the Royal Society of London, Series B: Biological Sciences 112(1933): 428-50.

a) 0.62$%b) 0.80$%c) 1%

02:47

Wen Z.

01:55

Amrita B.

04:08

Carson M.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 4

Indeterminate Forms and l'Hospital's Rule

Derivatives

Differentiation

Volume

Missouri State University

Harvey Mudd College

Baylor University

Boston College

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alright for this problem at least in part a all we have to do is write the equation and put in T Equals .05. And um r. Is three. So um he this is the equation equals I don't think I'd like to be using just regular arts. I'm gonna call this our one P equals one minus 10 to the negative P. Are one squared power Over P. R. one squared times. Natural log of 10. So The question is, what's the% 62%. So if we multiply by 62 I mean by 100 that'll give the percent. Exactly. Okay what's the percent luminous luminescence for a radius of two? Well then all I have to do is change the radius here to to And it's gonna be 80%. Does it make sense that it's larger? Well I would think that a two millimeter radius is going to need more light. Uh So no it doesn't make sense to me. Um Just like they observed that the amount of luminant since was not proportional the area, the pupil. It's it's inverse. That seems counterintuitive. Um What's the limit as R. Goes to zero from the plus side of P. So this is P when r goes to zero we're going to get P equals Um r goes to zero. Well as R goes to zero, the denominator becomes zero and then that's not good. So I'm going to take the derivative of the numerator. And the denominator, the derivative of the numerator is um -10. So let me write this down. It's going to be this copy that put that here. So that's and I'm still trying to just do the numerator. So let me see if I oops that didn't work. Okay. Wanted to put that negative up here. So it'll be itself times the derivative of the inside with respect to our which would be negative P. Um negative two P. Are one. Now I need to think about something here. Alright, since this was based 10, I also have to multiply by the natural log of 10. Okay so um that's what I get except the negative goes away. So let's think about this again. Oh bring this down. So it's going to be negative 10 to the negative P. R. One squared times the natural log of 10. And then I need to multiply that by the derivative of the inside which would be negative to pr one. So a negative times negative is a positive. Good. Got it. Now the denominator is going to be two p. Are one. Ah Mhm. Two P. Our one good Times The Natural Log of 10. It's a natural log of 10 cancels out. Um But nevertheless now can we get R. equals zero into there and no we can't. Okay. Um So let's simplify this to natural log of 10 times are one. Our times P cancels out in both the numerator and the denominator. So now I'm gonna have to take the derivative again. Uh This is not P but it's a derivative of peace. So I'm just gonna do that. Okay so now I'm going to take the derivative again. The derivative of the denominator is one. The derivative of the numerator is the derivative of the first. Which is one. I'm the second. I'm just gonna copy this, driven to the first times a second plus fruit of the second times the first. So that would be our one. And that's the first times the derivative of the second. Which would be that Times natural log of 10. So I'm gonna put that in here. Okay. Times the dream of the inside. So that's going to turn this to a negative and that's going to make this um driven the inside would be to pr that's going to give me an R squared on the outside. Okay, now I can put are one equals zero into this. So 10 To the zero power is one minus to pr one is zero. Uh times natural log of 10 times one. Okay, it's one. And so makes sense. Yes because the derivative or the limit uh the limitless r. Goes to zero is one. Which would be 100%. So the okay as the Hole size goes to zero, the luminescence tends towards 100%. So considering what I've learned then. Yes. I would expect it to go toward 100%. Thank you for watching

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