1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Geometric optics

  1. May 2, 2015 #1
    1. The problem statement, all variables and given/known data
    A fish 2cm long is floating in a spherical glass fishtank with radius 20cm. The glass is 0.8cm thick and has index of refraction n=1.56. The index of refraction of water is 1.33. Find the apparent position and length of the fish.

    2. Relevant equations


    3. The attempt at a solution
    I tried just applying the same equations I would normally:
    ##\frac{n_a}{s} + \frac{n_b}{s'} = \frac{n_b-n_a}{R}##
    s is object distance, s' image distance, R radius of curvature. I used s=10cm, s' unknown, R=-10cm and got s' to be -156/11.
    The magnification at this point is
    ##m=-\frac{n_as'}{n_bs}##
    So m= 1.209... and the fish appears to be 2.42cm long.
    Problem is, if this is the right approach I don't really know how to interpret the answers. Is the observer at the interface between materials? Where does 0.8cm come into it? Can I just apply the same equations again to find what the person outside the fishtank sees?
     
  2. jcsd
  3. May 2, 2015 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I know it's more work, but it helps readers a lot if you post all your working. In the above, I can't tell what you are plugging in for the indices without trying to reproduce your answer.
     
  4. May 3, 2015 #3
    Well, I was only considering the first boundary between the water and the glass, so I used na=1.33 and nb=1.56.Then I don't know what to do about the second boundary between the glass and air.
     
  5. May 3, 2015 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Since the glass is only a shell, it would probably more accurate to ignore the glass and just consider air and water.
    To bring the glass into it, draw a diagram. You have the fish on one side, a curved piece of glass, viewer on the other side. What are the radii of curvature of the inside and outside of the glass? Regard it as a lens.
     
  6. May 4, 2015 #5
    Does a lens of constant thickness behave according to the lensmakers equation? I think I can take the lens to be thin. We haven't actually covered lenses though...
     
  7. May 4, 2015 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes, I don't see why the glass would not behave approximately as a lens. You know the radii.
    But I hadn't noticed you have treated the water as a flat surface. That will behave as a lens too. Puzzled that you'd be given this question before covering lenses in your studies.
     
  8. May 5, 2015 #7
    Oh. I didn't know that about the water. I think I'll wait until my professor goes through the answers! Thank you for helping :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Geometric optics
  1. Geometrical Optics (Replies: 1)

  2. Geometrical Optics (Replies: 0)

  3. Geometrical optics (Replies: 5)

  4. Geometric Optics (Replies: 4)

  5. Geometric Optics (Replies: 3)

Loading...