Need help with physics and optics - lens power for myopia
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Need help with physics and optics - lens power for myopia

[From: ] [author: ] [Date: 12-04-27] [Hit: ]
But also how do compute where the new near point is??Thanks!f = -20 cm = -0.P = 1.(-.......
A very myopic person has a near point of 8 cm, and a far point of 20 cm. What power lens does he need to see clearly at a distance? What is his new near point with this vision correction?
I know I have to use 1/f = 1/s + 1/s' but I don't know what all to plug in for them. I know too that P = 1/f. But also how do compute where the new near point is??
Thanks!

-
s = ∞
s' = -20 cm

1/f = 1/s + 1/s' = 0 + 1/(-20)

f = -20 cm = -0.20 m

P = 1.(-.20) = -5 diopters
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