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Getletally belleved that nenflghludress ahect Jbcul 15"' of chdren certiin reglon A octool disuicl Iests Ine Vision Te4 ncoma undelgenan baxa1' How m...

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Getletally belleved that nenflghludress ahect Jbcul 15"' of chdren certiin reglon A octool disuicl Iests Ine Vision Te4 ncoma undelgenan baxa1' How mnany molld be expectad na-tsigtled? Whal E Iha Elandard duvllian fot the nunbat ol nenrn ghted ctkhon In thla go1p? nould bo oxoecled that Iha toxod chilcon _ namalghlod (Round Mo 4ecimal elacet naaded ,Mandord duvation I aoout(Rourd Io wo decimul plucus #4 nutded |

getletally belleved that nenflghludress ahect Jbcul 15"' of chdren certiin reglon A octool disuicl Iests Ine Vision Te4 ncoma undelgenan baxa1' How mnany molld be expectad na-tsigtled? Whal E Iha Elandard duvllian fot the nunbat ol nenrn ghted ctkhon In thla go1p? nould bo oxoecled that Iha toxod chilcon _ namalghlod (Round Mo 4ecimal elacet naaded , Mandord duvation I aoout (Rourd Io wo decimul plucus #4 nutded |



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The shiny surface of a $\mathrm{CD}$ is imprinted with millions of tiny pits, arranged in a pattern of thousands of essentially concentric circles that act like a reflection grating when light shines on them. You decide to determine the distance between those circles by aiming a laser pointer (with $\lambda=680 \mathrm{nm}$ ) perpendicular to the disk and measuring the diffraction pattern reflected onto a screen $1.5 \mathrm{m}$ from the disk. The central bright spot you expected to see is blocked by the laser pointer itself. You do find two other bright spots separated by $1.4 \mathrm{m},$ one on either side of the missing central spot. The rest of the pattern is apparently diffracted at angles too great to show on your screen. What is the distance between the circles on the CD's surface?

The geometry of the coalition and it can be predicted Officer managing behavior isn't? Which of the following? Is that correct? Yes, it is correct because magnetic behavior can be seen by the general first is the correct And I see him for two minutes involved the S two hi creation. Yes and I plus to have electronic conference and that is really 840 and seeing is that a strong field again that will pairing. So it will be the yesterday for all of us and it is within a group. Do you cannot value follows. It is incorrect because they don't know what they do for us five days then for me then P. D. C. LLC and for whom I am very scared. Ford is also correct. So correct. Option will be 1, 2 and four Palestine, correct.

This problem gets us to graph a circle with a radius of five. And we want to be able to see which windows will let us let us view it as an actual circle with that radius. And our conjecture from the last question was the X. Window must equal the Y. Window. So the only graph that should look like a perfect circle, there are the only window would be window three. So here they are, here is window one and you can see it's oval. So that matches our contract. Sure because the windows weren't equal window to also oval. This is the same um functions here. So kind of interesting how different they look. Window three. This is one we want to be a circle because their windows were equal to each other. So that's good news. And window for um doesn't give us a it doesn't give us a circle. Um It gives us more of an oval and its windows didn't match. So our conjecture that the X and window must equal the wire window for it to be a true shape is true.

Here for the solution. Let us consider that we is equal to invest. Distance U is equal to object Distance and f is equal to focal length. We know the equation that went upon we plus one upon you is equal to one point f Now when you is equal to infinity as light coming from infinity So here it will be one upon we plus one upon infinity is a cultivar upon f And from here we get we is equal to f Mhm for a mirror F is equal to our by two. So it will be equal to our by two. S V is equal to f has the distance between image and mirror is our by two? Yeah, Now for our minimum. So we have equal to seven, which is equal to our by two. So our which is to find our minimum will be seven multiplied by two which is equal to 14 now For our maximum, we have equal to 10. Mm. It is equal to our by two so are equal to our maximum which is equal to 20 ever. Uh huh Has the final answer is 14 27. So does the solution step by step. Please go through this. Thank you

We need to find the following value in radiance, so before you put it into your calculator, make sure your calculators in radiant mode. If not, you'll get the wrong answer. So when we plug this into the calculator, we get 1.2661 and this is in radiance.


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