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Focal Length of a Half Ball Lens - Physics Forums

Aug. 11, 2025

Focal Length of a Half Ball Lens - Physics Forums

Any help much appreciated - this is driving me crazy!

Homework Statement



Show that the focal length of a half ball lens, in air, (in the paraxial limit) is given by:
f=R/n-1


Homework Equations


Equation (from Hecht- Optics), for refraction at a spherical interface:

nm/s0 + nl/s1 = (nl-nm)/R

where,
nm = refractive index of medium
nl = refractive index of lens material
s0 = distance from object to surface
s1 = distance from image to surface

The Attempt at a Solution



Setting s0 = ∞
Then s1 = f

Subbing into the equation gives,
nm/s0 + nl/s1 = (nl-nm)/R

0 + nl/f = (nl-nm)/R

f = Rnl/nl-1


Where am I going wrong with this?
Thanks. OK... I think I figured this out. I am only taking the refraction from one spherical surface. I am treating it as if the rays are focusing inside the lens.

If I take the result of the first calc i.e. f=Rnl/nl-nm
and use it as the object distance for the second surface.
And bear in mind that the nl and nm will switch places for the second surface calculation, then I get the correct answer:

nl*nl-nm/Rnl + nm/s1 = nm-nl/R

nl-nm/R = nm/s1

f= s1 = nmR/nl-nm

Sorry if I wasted anyone's time on this. I literally had a brainwave, just after I wrote the post. Hello, HIB. Welcome to PF.

I'm not sure how you can take the focal length of the first surface as the object distance of the second surface. Wouldn't the object distance for the second surface be the distance from the image of the first surface to the second surface?

Also, when dealing with thick lenses, you have to define "focal length" carefully. Is it the distance of the image (of initially parallel rays) as measured from one of the surfaces of the lens? If so, which surface? Or is it measured from some reference point within the lens? I have the first edition of Hecht and Zajac's text. In that edition you can find quite a bit about different focal length definitions and how to calculate them for thick lenses in chapter 6.

You might also see what you get if you let parallel rays enter the lens on the flat side. Thanks TSny. You have a point about not using the focal length as the object distance.

If I approach it from the other side i.e. flat surface first, followed by half sphere surface.
Rays are approaching parallel, so s0=∞, surface is flat, so R = ∞

nm/∞ + nl/s1 = nl-nm/∞

nl/s1=0
→s1=∞

This makes sense, as the parallel rays will be undeviated after going through the flat surface.

So, then, I am back to treating just the half sphere surface, with the following params:
s0=∞; R is negative; nm and nl trade places in the equation as nm is now associated with the image and nl with the object

nl/∞ + nm/s1 = nm-nl/-R

s1= f = -R nm/ nm-nl
= R / (nl-1)

where nm=1

Correct? Now this question really startled me. We all know that from simple energy conservation, the ball can reach a height of 2l, i.e reach the top point of the vertical circle if a speed of is given at the bottom ##\sqrt{4gl}## as mentioned in the question. Hence, I expected the answer to be A. However the key says that the answer is D. Very confused, help is appreciated.

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