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!

Radius of curvature of glass and water lens

  1. Aug 19, 2014 #1
    1. The problem statement, all variables and given/known data

    by taking the lower curvature as r1 , and the upper curvature as r2 ,
    i dont know whether r1 is 20cm , r2 is 10cm or vice versa.

    But according to the ans r1= 10 cm , r2= 20cm . why is it so?

    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Aug 19, 2014 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Draw two circles the same size, with centres only a little apart. look at where the circles meet in relation to the two centres. The line joining the the intersections bisects the line joining the centres.
    If the circles have different radii then the intersections will be closer to one of the circle centres - which one?
     
  4. Aug 19, 2014 #3
    i drew the diagram , and it show what you have said. but i still dont understand why it is so . can you explain further?
     
  5. Aug 19, 2014 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Let me try another approach.
    Wrt the original diagram, the question is whether r1 can be the larger radius.
    Start with two circles of different radii, with the smaller inside and touching the larger. For reference, take the circle centres to be on the same X axis, with the point of contact on the left.
    If we shift the smaller to the left it will be partly outside, and the left half of the arrangement will look like the original diagram but with r1 as the smaller radius. So see what happens if we shift it to the right.
    At first, the smaller will be wholly inside the larger, no intersection. When intersection occurs again, the right half will again look like the original diagram, but again, with r1 as the smaller radius.
    If we continue moving the smaller to the right, until it is nearly outside the larger, neither the left half nor the right half of the arrangement looks quite like the original diagram. In each case, the lens stretches more than half way around the circles.

    You may think that's a lot to have to go through in order to solve the given problem. I didn't have to go through all that to solove it, because I just looked at the diagram and saw that r1 was the tighter curve, so must have the smaller radius. But I couldn't see a way to turn that into a solid argument.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted