A person’s right eye can see objects clearly only if they are between 25 cm and 75 cm away. (a) What power of contact lens is required so that objects far away are sharp? (b) What will be the near point with the lens in place
Focal length
1f=1u+1v\frac{1}{f}=\frac{1}{u}+\frac{1}{v}f1=u1+v1
1f=125−175\frac{1}{f}=\frac{1}{25}-\frac{1}{75}f1=251−751
1u=3−175\frac{1}{u}=\frac{3-1}{75}u1=753−1
u=37.5cmu=37.5cmu=37.5cm
Power
P=1FP=\frac{1}{F}P=F1
P=10.0735=2.67DP=\frac{1}{0.0735}=2.67DP=0.07351=2.67D
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment