A severely myopic patient has a far point of 10.00

keche0b 2021-12-14 Answered
A severely myopic patient has a far point of 10.00 cm. By how many diopters should the power of his eye be reduced in laser vision correction to obtain normal distant vision for him?
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Expert Answer

Linda Birchfield
Answered 2021-12-15 Author has 39 answers
Step 1
Given:
Far point =10cm
Step 2
The power of the normal eye is given by
P=(1f)=(1u)+(1v)
u=object distance=far point of a normal human eye is infinity
v=image distance=the image is formed on the retina which is about 2 cm from the lens
P=(1)+(10.02)=50D
The power of the myopic eye is given by
P=(1f)=(1u)+(1v)
u=object distance=far point of the myopic eye=10cm
v=image distance=the image is formed on the retina which is about 2 cm from the lens
P=(10.1)+(10.02)=60D
Reduction in power =6050=10D
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temnimam2
Answered 2021-12-16 Author has 36 answers
Step 1
The power of severely myopic eye pmyoπc can be obtained,
Pmyoπc=1d0+1di
Step 2
The quantity of dioptres pcorrected is,
Pcorrected=pmyoπcpal
Step 3
Thus, Pmyoπc=110.00×102m+12.00×102m=60.0D
Step 4
The power for the normal is 50D thus the corrected power is
pcorrected=60D50D
=10D
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