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!
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
s' = -20 cm
1/f = 1/s + 1/s' = 0 + 1/(-20)
f = -20 cm = -0.20 m
P = 1.(-.20) = -5 diopters