Focal Length of a Half Ball Lens

Click For Summary
SUMMARY

The focal length of a half ball lens in air, under the paraxial approximation, is defined by the formula f = R/(nl - 1), where R is the radius of curvature and nl is the refractive index of the lens material. The discussion highlights the derivation of this formula using the refraction equation from Hecht's "Optics," specifically nm/s0 + nl/s1 = (nl - nm)/R. The participants clarify the treatment of object distances for multiple surfaces and the implications of using different surfaces for focal length calculations.

PREREQUISITES
  • Understanding of paraxial optics and lens equations
  • Familiarity with the refraction equation from Hecht's "Optics"
  • Knowledge of refractive indices and their significance in optics
  • Basic principles of ray tracing in lens systems
NEXT STEPS
  • Study the derivation of the lens maker's equation for different lens shapes
  • Explore the concept of effective focal length in thick lenses
  • Learn about ray tracing techniques for complex optical systems
  • Investigate the impact of refractive index variations on lens performance
USEFUL FOR

Students of optics, optical engineers, and anyone involved in the design or analysis of lens systems will benefit from this discussion.

HIB
Messages
3
Reaction score
0
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.
 
Physics news on Phys.org
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?
 
Yes, that looks correct. Good!
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
2
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
4
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K