Optics B/C

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Re: Optics B/C

Post by blue cobra » February 5th, 2012, 10:35 am

I'm having trouble discovering how to answer questions involving eyeglasses. For example, let's say
A farsighted person has a near point 48cm away. If her glasses allow her to read a newspaper 23cm away, what is the power in diopters of her glasses?
I know that the power in diopters equals 1/f, so I tried treating this as a 2 lens system with the focal length of the eye at 48cm and the focal length of the system at 23cm, but I do the physical optics portion so I didn't get too far.
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Re: Optics B/C

Post by SciBomb97 » February 5th, 2012, 11:22 am

I bet forever knows the answer to that one.
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Re: Optics B/C

Post by Schrodingerscat » February 5th, 2012, 12:20 pm

blue cobra wrote:I'm having trouble discovering how to answer questions involving eyeglasses. For example, let's say
A farsighted person has a near point 48cm away. If her glasses allow her to read a newspaper 23cm away, what is the power in diopters of her glasses?
I know that the power in diopters equals 1/f, so I tried treating this as a 2 lens system with the focal length of the eye at 48cm and the focal length of the system at 23cm, but I do the physical optics portion so I didn't get too far.
http://www.wolframalpha.com/entities/ca ... /28/t7/d3/

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Re: Optics B/C

Post by foreverphysics » February 5th, 2012, 2:01 pm

Approximately 2.0869565217391204 diopters. I think.
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Re: Optics B/C

Post by cngu23 » February 6th, 2012, 4:36 am

Would the previous solution also apply for nearsighted people?
Or would there be a different equation, due to different lenses.
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Re: Optics B/C

Post by Schrodingerscat » February 6th, 2012, 5:06 am

If you change the search, you will also find the equation for nearsighted correction.

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Re: Optics B/C

Post by SciBomb97 » February 6th, 2012, 6:48 am

It has to do with the powers of the lenses of the glasses, in diopters.
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Re: Optics B/C

Post by foreverphysics » February 6th, 2012, 6:55 am

Solve it the same way you solved the hyperopia problem--the only difference is that the sign conventions should be different and the answer should be negative.

For example:
A nearsighted person cannot see objects clearly when they are beyond 50.00 com (the far point of the eye). What should the focal length of the prescribed lens be to correct this problem? What is the power of this lens (in diopters)?

You guys try solving this. I'll give you the answer once a few people have tried.
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Re: Optics B/C

Post by cngu23 » February 6th, 2012, 3:48 pm

foreverphysics wrote:Solve it the same way you solved the hyperopia problem--the only difference is that the sign conventions should be different and the answer should be negative.

For example:
A nearsighted person cannot see objects clearly when they are beyond 50.00 com (the far point of the eye). What should the focal length of the prescribed lens be to correct this problem? What is the power of this lens (in diopters)?

You guys try solving this. I'll give you the answer once a few people have tried.
According to the formula -1/max distance = lens power, it should be -2 diopters, which means a focal length of -0.5 meters?
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Re: Optics B/C

Post by blue cobra » February 7th, 2012, 4:19 pm

The Wolfram page repeatedly refuses to load on my computer. Would someone be so kind as to fill me in?
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