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This is problem 68 of chapter 23.
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In this problem we are asked to describe why is it that optical devices, so optical devices using visible light are not able to resolve objects at atomic scales or why we can't use visible light to resolve atoms.
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So let's start with the equation of aries disk.
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So here we have sine of theta equal to 1 .22 lambda over d.
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So let's take a moment to appreciate this equation.
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Here we have sign of theta, where theta is the angle from the central maximum to the first dark band, or that disk surrounding the resolution of light coming from a hole of diameter d, using a layer.
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Wavelength lambda.
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So if we are looking at objects on the size of atoms or atoms themselves, let's treat the atom itself as a source of light coming to our device.
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So if we treat the atoms themselves as our source of light or our aperture, we can say that d, capital d is on the order of one angstrom.
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So one angstrom, which is equal to 10 to the negative 10 meters.
01:47
So angstrom is a unit used for interatomic distances so this is the appropriate unit here for for this problem for distances between atoms or atoms themselves so we're gonna use approximately 10 to the 10 negative 10 here or or this capital d for lambda we are limited to 380 nanometers for the smaller wavelengths and then we're going to call this lambda viz because this is the limit for visible light 740 nanometers.
02:26
So notice that the wavelength is on the order of nanometers, whereas the denominator is going to be on the order of 10 to the negative 10.
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So let's see what happens if we pick an average wavelength here.
02:38
So let's pick an average wavelength.
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Well, let's be optimistic and let's try to choose the smallest wavelength that we can.
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But you're free to choose any wavelength here.
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The result is still going to be the same.
02:53
Let's go ahead and calculate sine of theta.
02:57
Let's suppose we go on the lower end.
03:00
Here we're using violet or maybe even ultraviolet light here.
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1 .22 times 308.
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Let's go to meters.
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So this is going to be times 10 to the negative 9 meters.
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Here in the numerator, here in the denominator, we're going to have 10 to the negative 10.
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Meters.
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Let's see what this comes out to be...