How Does Refractive Power Affect Vision Clarity at Different Distances?

In summary, the student is trying to find the distance between her eyes and the blackboard, given the refractive power of her contact lenses and the distance between the lens and her retina. She then uses the equation 1/f = 1/di + 1/do to calculate the distance of the object from the lens. In part (b), the student uses the distance of the object, the distance of the image, and the height of the object to find the size of the image on her retina. However, she confuses the variables do and di in her calculations.
  • #1
cheechnchong
132
1

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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  • #2
(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?
 
  • #3
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??
 
  • #4
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!)
 

What is "Contact Optics"?

Contact Optics is a branch of optics that deals with the study of how light behaves when it comes into contact with different materials, such as lenses, mirrors, and other optical components.

What are the main applications of Contact Optics?

Contact Optics has a wide range of applications, including designing and manufacturing lenses for glasses and cameras, creating optical instruments such as microscopes and telescopes, and developing advanced optical systems for medical imaging and communication technologies.

How does Contact Optics differ from other branches of optics?

Contact Optics focuses specifically on the interaction between light and materials, while other branches of optics, such as geometric optics and physical optics, deal with different aspects of light, such as its propagation and wave behavior.

What are some key concepts in Contact Optics?

Some key concepts in Contact Optics include refraction (the bending of light as it passes through different materials), reflection (the bouncing of light off a surface), and absorption (the process by which materials absorb and convert light energy).

How is Contact Optics used in everyday life?

Contact Optics plays a crucial role in our daily lives, from correcting vision with glasses or contact lenses to capturing images with cameras and using laser technology for various purposes. It is also used in industries such as telecommunications, transportation, and manufacturing.

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