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.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?
Optics B/C
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Re: Optics B/C
I'm having trouble discovering how to answer questions involving eyeglasses. For example, let's say
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Re: Optics B/C
I bet forever knows the answer to that one.
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Re: Optics B/C
http://www.wolframalpha.com/entities/ca ... /28/t7/d3/blue cobra wrote:I'm having trouble discovering how to answer questions involving eyeglasses. For example, let's say
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.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?
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Re: Optics B/C
Would the previous solution also apply for nearsighted people?
Or would there be a different equation, due to different lenses.
Or would there be a different equation, due to different lenses.
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Re: Optics B/C
If you change the search, you will also find the equation for nearsighted correction.
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Re: Optics B/C
It has to do with the powers of the lenses of the glasses, in diopters.
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"So whether you eat or drink or whatever you do, do it all for the glory of God." ~1 Corinthians 10:31~
They say that a smile can light up somebody's day
So today, smile
Shine a light in somebody's life
Be that light in the darkness
"So whether you eat or drink or whatever you do, do it all for the glory of God." ~1 Corinthians 10:31~
They say that a smile can light up somebody's day
So today, smile
Shine a light in somebody's life
Be that light in the darkness
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Re: Optics B/C
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.
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
According to the formula -1/max distance = lens power, it should be -2 diopters, which means a focal length of -0.5 meters?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.
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11 MD Regional
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Walter Johnson 09-11
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Material S (2/-) TPS (-/2)
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Remote S(2/6) Water Q(2/4) Optics (-/5)
11 MD Regional
Remote S(2) Eco(2) D Planet(3)
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Re: Optics B/C
The Wolfram page repeatedly refuses to load on my computer. Would someone be so kind as to fill me in?
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