UIUC Phys 102: Exam 3 Prep - Questions 4-6

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http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam3/sp06

Questions 4-6 , i don't need it to be solved, i just need to know whether my equations are right cause i have someone else's answers and i don't know how he got them. he Has I1 = I0 / 2 which doesn't make sense

I1 = I0 Cos(45)^2 since it is not someone correct? if it was verticel it would be I1 = I0/2

I2 = I1 (Cos 45 + Theta ) ^2 since u are trying to find the angle diffrence

I3 = I2 (Cos 45 - Theta) ^2

is that even right at all? if not does anyone know of a site that has lectures to this stuff. Thanks

 

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http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam3/sp06

Question 19. How do u find what order fringe the 2nd and 3rd are in that pic?

 

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http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam3/sp05

Question 8

So the first one since it is horizontal would be I1 = I0/ 2 so this is only true if it is horizontal or vertical

 

[Max Eilerson]

http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam3/sp06

Question 19. How do u find what order fringe the 2nd and 3rd are in that pic?

To get the order of the fringe you just count from the centre, there is no zeroth order for a dark fringe the first dark fringe is the first minimum (trough), the second the second minimuetm etc (from the right of the central line). Anyway that's irrelevant the distance between the 2nd and 3rd fringe is just the distance between the central maximum (0 order) and the first maximum (n=1), which you should know how to find.

 

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To get the order of the fringe you just count from the centre, there is no zeroth order for a dark fringe the first dark fringe is the first minimum (trough), the second the second minimuetm etc (from the right of the central line). Anyway that's irrelevant the distance between the 2nd and 3rd fringe is just the distance between the central maximum (0 order) and the first maximum (n=1), which you should know how to find.

more info please :)

Edit: Actually this is what i did


Sin Theta = ( 2.5)(600*10^-9) / ( .3*10^-3) = .005
Sin Theta = (3.5)(600 * 10^-9) / (.3*10^-3) = .007

Angle is .28648
Angle is .40107

Y = Tan theta * L

Did it for both of them, subtracted and i got 1 mm, is the way i solved it correct or would u have had to use 1.5 and 2.5? thanks

 

[Max Eilerson]

Yes this answer is fine, although your working could be a lot of quicker and neater. Do you know what a small angle approximation is? As \theta tends to 0, sin \theta tends to \theta.ame with tan, cos tends to 1 of course.

 

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http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam3/fa05

Question 6

So i found the focal length and it is 12.5 cm which happens to be the answer. now i just want to make sure that it will be true for other problems if the magnification is half the size. So if it said quarter of a size would it be 6.25 cm instead? Thanks

 

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http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam3/sp04
QUestion 2
is this something i would have to just know or can i get it form I/d + 1/ O = 1 /F

 

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http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam3/sp04
Question 6
What would be the answer? It is C but i want to know how to solve it. thanks

 

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All right, i got all the stuff up there, i just need help on this one

http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam3/fa04

Why is A wrong?

So if the focal Length is 5 for the convex then they were converge there

 

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Ignore everything that is above me



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QUestion 25 and 26