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Homework Help: Geometry and the principles of a spherometer

  1. May 13, 2014 #1

    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?


    Attached Files:

  2. jcsd
  3. May 13, 2014 #2


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    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


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    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.
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