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Find the power of the corrective lens needed to bring his near point to 25.0cm

2007-04-04 03:36:58 · 2 answers · asked by rkh 1 in Science & Mathematics Physics

2 answers

The vergence corresponding to that distance is 1/0.6 =1.66Dt

to see et 25cm =0.25m, you need a vergence of 1/0.25 =4Dt

so the power of the lense is 4-1.67 = 2.33Dt

Please : I use the notation in my country

We use vergence instead of power. I hope it will nethertheless be useful

2007-04-04 04:07:25 · answer #1 · answered by maussy 7 · 0 0

1

2016-06-20 00:41:55 · answer #2 · answered by ? 3 · 0 0

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