Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 25 - Optical Instruments - Problems - Page 740: 21

Answer

Her new near point is 18.4 cm and her new far point is 100 cm.

Work Step by Step

Use the lens equation. Calculate the object distance for the contact lens to form an image at her eye’s old near point of 10.6 cm. $$P=\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$$ $$d_o=\frac{d_i}{Pd_i-1}=\frac{-0.106m}{(-4.00D)(-0.106m)-1}=0.184m=18.4cm$$ This is her new near point. Use the lens equation. In the same way, calculate the object distance for the contact lens to form an image at her eye’s old far point of 20.0 cm. $$d_o=\frac{d_i}{Pd_i-1}=\frac{-0.200m}{(-4.00D)(-0.200m)-1}=1.00m=100cm$$ This is her new far point.
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