How Does Refractive Power Affect Vision Clarity at Different Distances?

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SUMMARY

The discussion focuses on calculating the distance from a student's eyes to a blackboard using the lens formula, where the refractive power of the contact lenses is 57.50 diopters and the lens-to-retina distance is 1.750 cm. The focal length (f) is determined to be 0.0174 m, leading to an object distance (do) of 0.328 m. For the image size on the retina, the student calculates the height of the image (hi) as -0.945 m using the magnification formula, but there is confusion regarding the correct application of object distance (do) and image distance (di).

PREREQUISITES
  • Understanding of lens formulas, specifically 1/f = 1/di + 1/do
  • Knowledge of refractive power and its calculation in diopters
  • Familiarity with magnification equations, including m = hi/ho
  • Basic principles of optics related to image formation
NEXT STEPS
  • Study the derivation and application of the lens formula in optics
  • Learn about the significance of positive and negative signs in optics calculations
  • Explore the concept of magnification and its implications in real-world scenarios
  • Investigate the effects of different refractive powers on vision clarity
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Students studying optics, physics educators, and anyone interested in understanding the relationship between refractive power and image formation in lenses.

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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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