How Does Refractive Power Affect Vision Clarity at Different Distances?

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Homework Help Overview

The discussion revolves around a physics problem involving optics, specifically focusing on the concepts of refractive power, object distance, and image size in relation to lenses. The original poster presents a scenario where a student is using contact lenses with a specified refractive power to read from a blackboard, leading to questions about the distances involved and the resulting image size on the retina.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of object distance (do) and image distance (di) using the lens formula. There is confusion regarding the signs and definitions of do and di, as well as the application of magnification to find the image height (hi).

Discussion Status

Some participants provide guidance on identifying the correct distances and equations to use, while others express uncertainty about the calculations and terminology. Multiple interpretations of the problem setup are being explored, particularly regarding the relationships between the distances and the resulting image size.

Contextual Notes

Participants note the potential confusion arising from the signs in the lens formula and the definitions of object and image distances. There is also mention of the original poster's uncertainty about their calculations and the need for clarification on the application of magnification.

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Homework Statement



A student is reading a lecture written on a blackboard. Contact lenses in her eyes have a refractive power of 57.50 diopters; the lens-to-retina distance is 1.750 cm. (a) How far (in meters) is the blackboard from her eyes? (b) If the writing on the blackboard is 5.00 cm high, what is the size of the image on her retina?

Homework Equations



1/f = 1/di + 1/do


The Attempt at a Solution



(a) I know that once you get f (meters) using the given diopters measurement, you can use that to get an image. f = .0174 m...and then i get di = 0.328 m. (b) i substract 5 cm from 1.75 cm to get object distance...then i use f = .0174 m...finally, i plug in the equation and get di = 27 m.

I am not sure if I did this right because the negative and positive stuff really bother me. anyways, help me out if you see something wrong and let me know if i "might" right too! thanks.
 
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(a) Here you have to find do, the distance between the object and the lens. di is given so once you have f you just use you're equation to get do. Are you confusing do and di? Other than that it looks like you did it right (i didn't check the math though)
(b) I'm not sure what you did here. You have do, di and ho already, what equation can you use to get hi?
 
daniel_i_l said:
(a) Here you have to find do, the distance between the object and the lens. di is given so once you have f you just use you're equation to get do. Are you confusing do and di? Other than that it looks like you did it right (i didn't check the math though)
(b) I'm not sure what you did here. You have do, di and ho already, what equation can you use to get hi?

hey i appreciate the help...

i resolved (b) using the do, di, and ho

m = - (0.33)/ (.0175) = -18.9 m

and then i plug it into m = hi/ho

hi = (-18.9 m)(.05 m) = -0.945 m

do you think i approached this part right??
 
Almost!
But it's a little confusing because you're using di for the distance of the object which is really do. So when you wrote - (0.33)/ (.0175) that really is
-do/di and it should be -di/do (di = 0.175 not do!)
 

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