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Problem 25.67Part AHowaway can human eye dlstingulsh two car headlights 2.1 pan? Conslder only dlffracilion effects and assume an eye pupil dlameter of 6.5 tntr and...

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Problem 25.67Part AHowaway can human eye dlstingulsh two car headlights 2.1 pan? Conslder only dlffracilion effects and assume an eye pupil dlameter of 6.5 tntr and wavelength of 560 nm Express your answer t0 two signlficant flgures and Include the appropriate unlts:ValueUnitsSubmltRequest AnswerFartWhat the minimum argular separation an oye could resolve when viewing two stars , considering only diffraction effects? Express your answee using two significant figures.AZWaf arcSubmitRequest Angwer

Problem 25.67 Part A How away can human eye dlstingulsh two car headlights 2.1 pan? Conslder only dlffracilion effects and assume an eye pupil dlameter of 6.5 tntr and wavelength of 560 nm Express your answer t0 two signlficant flgures and Include the appropriate unlts: Value Units Submlt Request Answer Fart What the minimum argular separation an oye could resolve when viewing two stars , considering only diffraction effects? Express your answee using two significant figures. AZW af arc Submit Request Angwer Part ' This quostion will be shown after you complete previous question(s). Provide Feedback



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(II) $(a)$ How far away can a human eye distinguish two car headlights 2.0 $\mathrm{m}$ apart? Consider only diffraction effects and assume an eye pupil diameter of 5.0 $\mathrm{mm}$ and a wavelength of 550 $\mathrm{nm}$ . (b) What is the minimum angular separation an eye could resolve when viewing two stars, considering only diffaction effects? In reality, it is about $1^{\prime}$ of arc. Why is it not equal to your answer in $(b) ?$

It is very similar to example twenty five point eleven. So I'd hardly recommend you to look at this example on your textbook and with these in the same notation to solve the problem. So here will be using the same equation as this example and will be solving for and to find out how far we can see from here we saw for N that's going to be data over s. But actually, before doing that, we have to use the relay criteria which gives us the angle a separation, our angle resolution. So we'LL be using question twenty five point seven, which is the angler resolution so equal to one point two two times lambda over Mindy So we can substitute this guy over here, and ah then we'LL have l equals one point two lambda over d or we're s so that's Ah no, no, That's going to be the opposite S o l is going to be s over, Terra. So that's the Times s divided by one point Toto Landa Sorry, Loan debt. Now we're given all the valley So we're given D as six point zero times ten to the bar minus three meter then s is given us two meter. Then Lambda is five sixty in enemy ER, and we convert that a meter. So from that we get the the the distance as one point eight times ten to the bar for meter or eighteen kilometer. Okay, Uh, it is approximately eleven miles. All right, so that's part a now for part B. Ah, we use the same data for ah, the eye in the wave length. So that means data will be one point two times Landover. Maybe So this will be one point two two times five sixty them sent apart. Minus nine meter. Divided by the diamond of it is six point zero times ten to the bar, minus three meter. This's equal to one point one three nine. I'm Santa Bar minus four. Brilliance out. So now we need to convert that toe second or like degree second. So that degree, or in our case, it will be second because it's a very small value. Ah, but the second is not equal to the unit of time. This is the unit of degree. Our dessert. Here s o. Let's convert that. It's gonna be one point one between intense in the ominous for radiant times. We can without a degree first, so there's gonna be one eighty over five radiant. Then we convert that to second. So one degree is thirty six hundred seconds. So that's that's so if it's all for it, we get twenty three second as our answer. Now this is less than the actual resolution, which is giving us one a minute. It's on DH. That's because we have the atmospheric ethics and, ah, aberrations in the eye. So the value is less because at Morse Munich, it affects and emburey tions India.

Okay, so you're in chapter 35 from 71. So it says, how far the power part can the human eye distinguished to car headlights two meters apart. Says consider only two fraction of sex and assume. And I'd diameter of six millimeters and a wavelength of 5 60 mana meters. So this problem is really similar to a example 35 6 if you are back at that. But we're gonna set up this in a very similar way. Such that we have best is our distance. We can distinguish. And this is given by El Data, which is the same as the 1.2 to Lambda over Big D. So this big D here is actually this deeper trying to fight passes The separation between he's an el is the distance away that we could distinguish this. So what we do is we rearrange for l Mrs D s over 1.22 older. So we have six millimeters or six times 10 to the negative, three meters times two meters over 1.22 times 5 60 times 10 The negative 98 of this what we get here. 18 kilometers Cool. So Basically, if we look at the cars distance greater than eking kilometers away both their headlights, they're gonna look like one would like to us. So Part B says, What is the minimum angular separation? And I resolved when viewing two stars considering only diffraction events. In reality, it's about one set second there one minute of lark. Why is that? I decline to answer in part B capes over the same way we're trying to sell to the angle of separation. Here's 1.2 to Lambda over tea. So if we just club that in what we get is 1.139 times 10 to the negative four Ray deepens. So how do we convert that to a second? Well, first cover two degrees and then we convert Teoh seconds while there's 3600 seconds and one degree. So this comes out to being 23 exits. Cool. So this is actually less than the real resolution that we can see, which is about one minute with our eyes in this, because of all the atmospheric effects and aberrations in our I. So this is like the theoretical maximum of how could we could see stars, but that's not exactly how we actually see him

This problem covers the concept of a resolution angle and the angular separation for the regulation for the human eye is given by this situation where D. Is the diameter of stupid. So for party the anglo separation throughout our equals 1.22 into the wavelength emitted by the lights. That is 550 nm. Okay, Upon the diameter of the people, that is five. Okay, So the angular separation in uh radiance is 1.3 and two And there's -4 radiance. No party to find the Maximum distance where the two uh Lights can be resolved. Let's consider this as source one. And this has source too. And it is making angle theta to the eye and its distance from uh the observer is uh that's A D. And the separation between the two sources x. And if we use the small angle approximation that you can write theater equals X upon deep. Therefore the distance of the souls. The maximum distance tree is equivalent to ex upon data, I subscribed the value. So the maximum distance of the uh to light from the person is if you were into Access 1.4 m upon 1.3 into and there is a minus for Canadians, R. D. Equals 1.0 into. And it's four m as you can see the maximum distances 10 kilometers

For this problem on the topic of diffraction, we're told that an approaching car has headlights that at 1.4 m apart, we want to know the angular separation and the maximum distance at which the I will resolve them. And we are told that the pupil, the amateurs, five and the wavelength is 550 nm. Yeah, now, by the really criteria, two point sources can be resolved if the central diffraction maximum of one source is centered on the first minimum of the diffraction pattern of the other. Thus the angular separation of the sources must be at least peter are, which is 1.22 Lambda over D where land is the wavelength and the the amateur of the aperture. Now for the headlights in this problem We have to to our is able to 1.22 and were given The wavelength of light to be 550 Times 10 to the -9 meters, divided by five times 10 to the minus three meters, which gives 1.3 Times 10 to the -4 radiance. Mhm. So, and that's the angular separation at which the I will resolve them. Next we were to find the maximum distance. Now, if we let l be the distance on the headlights to the eye, when the headlights are just resolvable and big D the separation of the headlights, then the separation of the headlights D. Is equal to the distance from the headlights. L times Theta are. Now here we made the small angle approximation and so we can rearrange and solve for L. We get L two B. D over Vita are And this is 1.4 m divided by 1.34 Times 10 to the -4 radiance. And so the distance is one Times 10 to the power for m, which is 10 kilometers.


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