Geometry and the principles of a spherometer

[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.

Homework Statement

"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.

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.

Please can you help?

Thanks!

 

[HallsofIvy]

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}.

 

[BOAS]

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}.

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...

 

[LCKurtz]

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.

Homework Statement

"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.

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.

Please can you help?

Thanks!

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.

 

[BOAS]

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.

Thank you!

Got it.