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Radius of curvature of the cornea 
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#1
Nov2008, 09:25 PM

P: 24

1. The problem statement, all variables and given/known data
The cornea, a boundary between the air and the aqueous humor, has a 3.0cm focal length when acting alone. What is its radius of curvature? 2. Relevant equations n(1)/s + n(2)/s' = (n(2)n(1))/R where, n is the index of refraction (unitless), s' is the img dist (cm). and s is the obj. dist (cm). R is the radius (cm). also, 1/s + 1/s' = 1/f where, f is the focal length. also, n(1) is air having a index of refraction of 1.00 and n(2) is the aqueous humor having a index of refraction of 1.34 3. The attempt at a solution So, I assumed since only f was given as 3.0cm that 1/s' + 1/s could be 1/6.0cm + 1/6.0cm = 1/3.0cm. Then that would give 1.00/6.0cm + 1.34/6.0cm = (1.341.00)/R giving R to be 0.87cm but it says that is wrong. Any ideas??? Also, I thought there was a equation editor for this website? I can't seem to see it... Is it only for the premium members? 


#2
Nov2008, 09:45 PM

P: 195

Radius of curvature is the distance from the cornea to the centre of curvature. If the focal length is half this distance, what do you think the answer is?



#3
Nov2008, 10:30 PM

P: 24

Are you implying that it is 2*f=R? 6.0cm is not the right answer if so. That only applies for mirrors right? They seem to be indicating the "air" and the "aqueous humor" in the problem... both of which have given index of refractions in the book. I believe the equation used to solve this problem will deal with the index of refraction... It is hinted in the problem statement pretty much. Thanks for the help though... let me know if you have any other thoughts.



#4
Nov2108, 11:37 AM

HW Helper
P: 5,341

Radius of curvature of the cornea
As for your Equation editing there is a Σ at the top of the message window that is an aid to inserting LaTex into messages. If you want to see how some equations are constructed just click on them and the underlying LaTex is revealed.
You can also take advantage of the X^{2} and X_{2} buttons for subscripting and superscripting. n(1)/s + n(2)/s' = (n(2)n(1))/R yields: [tex] \frac{n_1}{s_1} + \frac{n_2}{s_2} = \frac{n_2  n_1}{R} [/tex] As to your problem ... Wouldn't the lens makers relationship be the way to approach it? [tex] \frac{1}{f} = (n1)*(\frac{1}{R} + \frac{1}{R}) [/tex] 


#5
Nov2208, 01:20 PM

P: 24

I guess you could approach it from the lens makers... I don't see how though. I'm given only one radius... in your formula that needs to be a R subnot 1 and 2. Also, I'm dealing with two index of refraction values. With the amount of known values I have I cannot seem to piece them together into a formula that satisfies utalizing them all. Am I missing something? Am I supposed to use only one index of refraction and pull another radius out of the sky? hmm, thanks for the help anyways... always appreciated. i'm just frustrated... it is due tomorrow morn.



#6
Nov2208, 02:03 PM

P: 166

It is useful to think about the definition of focal length. How is focal length defined?



#7
Nov2208, 03:01 PM

HW Helper
P: 5,341




#8
Nov2208, 03:06 PM

P: 166




#9
Nov2208, 07:38 PM

P: 24

I did f*(n1) = R even before you mentioned the equation in that form. Maybe the masteringphysics website is off. I derived that formula by taking s=f and s' being equal to inifinity for an answer of 1.02cm (we are on the same boat). Yes, it is focusing into a medium with an index of 1. This beats me. I'm down to two attempts left. I'll still pass this class just fine... I just hate not getting an answer.



#10
Nov2208, 08:22 PM

P: 166

Well, you're almost correct. What do s and s' refer to? Where is the light coming from, and where is it focusing (remember, we're talking about the eye, not an abstract mathematical equation )?



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