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Exam III Help

  1. Nov 6, 2006 #1
    http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam3/sp06 [Broken]

    Questions 4-6 , i dont need it to be solved, i just need to know whether my equations are right cause i have somone else's answers and i dont know how he got them. he Has I1 = I0 / 2 which doesnt make sense

    I1 = I0 Cos(45)^2 since it is not verticle 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
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Nov 6, 2006 #2
    http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam3/sp06 [Broken]

    Question 19. How do u find what order fringe the 2nd and 3rd are in that pic?
     
    Last edited by a moderator: May 2, 2017
  4. Nov 6, 2006 #3
    http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam3/sp05 [Broken]

    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 verticle
     
    Last edited by a moderator: May 2, 2017
  5. Nov 7, 2006 #4
    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.
     
    Last edited by a moderator: May 2, 2017
  6. Nov 7, 2006 #5
    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
     
    Last edited: Nov 7, 2006
  7. Nov 8, 2006 #6
    Yes this answer is fine, although your working could be alot 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.
     
  8. Nov 8, 2006 #7
    http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam3/fa05 [Broken]

    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
     
    Last edited by a moderator: May 2, 2017
  9. Nov 9, 2006 #8
    http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam3/sp04 [Broken]
    QUestion 2
    is this something i would have to just know or can i get it form I/d + 1/ O = 1 /F
     
    Last edited by a moderator: May 2, 2017
  10. Nov 9, 2006 #9
    http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam3/sp04 [Broken]
    Question 6
    What would be the answer? It is C but i want to know how to solve it. thanks
     
    Last edited by a moderator: May 2, 2017
  11. Nov 11, 2006 #10
    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 [Broken]

    Why is A wrong?

    So if the focal Length is 5 for the convex then they were converge there
     
    Last edited by a moderator: May 2, 2017
  12. Nov 12, 2006 #11
    Ignore everything that is above me



    http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam3/sp00 [Broken]

    QUestion 25 and 26
     
    Last edited by a moderator: May 2, 2017
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