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Physics 3: Images, Interference and Difraction

  1. Mar 19, 2014 #1
    So three problems im stuck on.
    1.) You have a converging lens with n=1.5,for a symmetric lens so the two lens have same magnitude, what should the radius of curvature be so the focal length is 10cm. I know 1/f = (n-1)[1/r1-1/r2] so that is 1/0.10m =(1.5-1)[1/r1-1/r2] where im lost is in the r1 and r2. would r2 be -r1 so we get 2r?

    2)A 3cm tall object is located 10cm in front of a converging lense with a focal length of 15cm. What is the magnification of the imaged formed? |m|= h'/h and m =-i/p h is 3cm and f=15cm and p=10cm which gives a i =-30cm so would the image m be 3cm? or am i missing something?

    3.) The minimum width of a slit for single diffraction to produce an interference pattern is? no min, lamda, lambda/2, or 2 lamda. since its sin=lambda/a where a is the slit width as long as a>0 theres no limit?
  2. jcsd
  3. Mar 20, 2014 #2

    Simon Bridge

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    1. both sides have the same radius - so there are only two possibilities.
    Try both and see which makes sense.

    2. check by sketching the ray diagram

    3. this one will depend on your course.
    strictly - there is no maximum, you are right. You also get diffraction for the limit of zero width.
    diffraction can occur around an edge - so the "slit width" there is infinite.
    however - your course may be talking about a specific kind of diffraction ... so check your notes.
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