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Resolution of the Eye. The maximum resolution of the eye depends on the diameter of the opening of the pupil (a diffraction effect) and the size of the retinal cells. The size of the retinal cells (about 5.0$\mu \mathrm{m}$ in diameter) limits the size of an object at the near point $(25 \mathrm{cm})$ of the eye to a height of about 50$\mu \mathrm{m} .$ To get a reasonable estimate without having to go through complicated calculations, we shall ignore the effect of the fluid in the eye.) (a) Given that the diameter of the human pupil is about 2.0 mm, does the Rayleigh criterion allow us to resolve a 50 -\mum-tall object at 25 $\mathrm{cm}$ from the eye with light of wavelength 550 $\mathrm{nm}$ ? (b) According to the Rayleigh criterion, what is the shortest object we could resolve at the $25-\mathrm{cm}$ near point with light of wavelength 550 $\mathrm{nm} ?(\mathrm{c})$ What angle would the object in part (b) subtend at the eye? Express your answer in minutes $\left(60 \min =1^{\circ}\right),$ and compare it with the experimental value of about 1 min. (d) Which effect ismore important in limiting the resolution of our eyes: diffraction or the size of the retinal cells?

a. 84b. 84$\mu \mathrm{m}$c. 1.155 $\mathrm{min}$d. that the effect of diffraction is more important in limiting the resolution of our eyes

Physics 103

Chapter 36

Diffraction

Reflection and Refraction of Light

Rutgers, The State University of New Jersey

Simon Fraser University

University of Winnipeg

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in part a were using two millimeters as the diameter of the retina of an eye and a wavelength of 550 nano meters. So we're used also using a distance to the object of 0.25 meters, um, and were asked if we can resolve a 50 micrometer tall object. So, um, the sign of Fada is 1.22 Lambda over D. And the tangent of Fada is X over L where X is the resolution. So X is L Tangent of Fatal, which is the inverse sign of 1.22 Lambda over D. So I'm going to put that into a calculator. L is 0.25 kinds Tangent of inverse sign of 1.22 times Lambda, which is 150 times 10 to the negative ninth power over DE, which is 0.2 That gives me X of 123456 um, 84 micrometers. I just want to make sure I didn't make any mistakes. 0.25 10 Jin inverse sign 1.22 Lambda, uh, over 0.2 Okay, so can we resolve a 50 micron tall object. The answer's no. You cannot, because 85 is the minimum. Be the shortest object we could resolve that looks like it's the same question again. It's just the 84 micrometers. See what bank? Okay, what angle would it sub tend The I? Well, the tangent of data is X over. L so Theda would be the inverse tangent of X over L. So I'm going to get that in degrees. Inverse tangent Set my calculator two degrees x over l l is 0.25 That gives me 0.19 degrees, but one degree is 60 minutes. So if I multiply that by 60 could give me minutes 1.2 minutes. OK, moving on Teoh Part de least. I think there's a part D Let's see here. So which is more limiting diffraction or the retinal cells? Well, the retinal cells have a real resident resolution of 50 micrometers and diffraction had a resolution of 84 micrometers so diffraction ISMM or limiting good 84 versus 50

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