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!

Geometry and the principles of a spherometer

  1. May 13, 2014 #1
    Hello,

    i'm struggling to understand the equation i've been given for finding the radius of a sphere by using a spherometer. I wasn't sure if this would be better in the physics section, but I figured it is essentially geometry.

    1. The problem statement, all variables and given/known data

    "From the diagram, simple geometry shows that the radius, r may be calculated from the formula [itex]r = \frac{h^{2} + l^{2}}{2h}[/itex]"

    See attached for the diagram.

    3. The attempt at a solution

    The [itex]h^{2} + l^{2}[/itex] term makes me think that the curved surface is being approximated as the hypotenuse of a right angled triangle, but I can't make sense of where dividing by [itex]2h[/itex] gets you.

    I've been trying to relate it to the formula for the radius of a circle using an arc, but i'm npt getting anywhere.

    Please can you help?

    Thanks!
     

    Attached Files:

  2. jcsd
  3. May 13, 2014 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    We can take the equation of the sphere to be [itex]x^2+ y^2+ z^2= r^2[/itex]. The circular boundary of the lens is at z= r- h so [itex]x^2+ y^2+ r^2- 2rh+ h^2= r^2[/itex] and then [itex]x^2+ y^2- 2rh+ h^2= 0[/itex] or [itex]x^2+ y^2= 2rh- h^2[/itex].

    That is a circle with radius [itex]l= \sqrt{2rh- h^2}[/itex].
     
  4. May 13, 2014 #3
    I don't understand exactly what you've done. What is the 'circular boundary of the lens'?

    I follow your steps algebraically, but the resulting formula is not the same as stated in my lab book. Are the two actually equivalent?

    I am confused what 'r' denotes in your explanation if you're using l for radius...
     
  5. May 13, 2014 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Draw the radius from the center to the spherometer leg. That forms a right triangle with hypotenuse ##r## and legs ##l## and ##r-h##. Use the Pythagorean theorem and solve for r.
     
  6. May 13, 2014 #5
    Thank you!

    Got it.
     
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: Geometry and the principles of a spherometer
  1. Geometry Problem (Replies: 4)

  2. Geometry, planes (Replies: 2)

  3. Coordinate geometry (Replies: 1)

  4. Basic geometry (Replies: 1)

  5. Geometry problem (Replies: 3)

Loading...