# Geometry and the principles of a spherometer

1. May 13, 2014

### BOAS

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 $r = \frac{h^{2} + l^{2}}{2h}$"

See attached for the diagram.

3. The attempt at a solution

The $h^{2} + l^{2}$ 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 $2h$ 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.

Thanks!

#### Attached Files:

• ###### spherometer.PNG
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2. May 13, 2014

### HallsofIvy

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

That is a circle with radius $l= \sqrt{2rh- h^2}$.

3. May 13, 2014

### BOAS

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...

4. May 13, 2014

### LCKurtz

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.

5. May 13, 2014

Thank you!

Got it.