A myopic adult has a far point at 0.1m. His power of accomodation is 4 diopters.
- What power lenses are required to see distant objects?
- What is his near point without glasses?
- What is his near point with glasses? (Take the image distance from the lens of the eye to the retina to be 2cm.)
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If two thin lenses of focal lenght $f_1$ and $f_2$ are in contact, then the effective focal length of the combination is given by
$\frac{1}{\text{f}}=\frac{1}{\text{f}_1}+\frac{1}{\text{f}_2}$
Also the effective power of the combination is given by
$P = P_1 + P_2$
Substituting $f_1 = 0.1$ and $\text{f}_2=\frac{1}{40}=0.02$, we get
$\text{P}_\text{f}=\frac{1}{\text{f}}=\frac{1}{0.1}+\frac{1}{0.02}=60\text{D}$
By the corrective lens the object distance at the far point is infinity.
The power required is given by
$\text{P}_\text{f}=\frac{1}{\text{f}'}=\frac{1}{\infty}+\frac{1}{0.02}=50\text{D}$
For eye + lens system, we have the sum of the eye and that of the lgasses $P_g$
$\therefore\ \text{P}_\text{f}=\text{P}_\text{f}+\text{P}_\text{g}$
$\Rightarrow\ \text{P}_\text{g}=\text{P}_\text{f}-\text{P}_\text{f}$
Substituting $P_f = 60D$ and $P'_f = 50D$, we get
$\text{P}_\text{g}=-10\text{D}$
$\therefore\ 4=\text{P}_\text{n}-\text{P}_\text{f}$
or $\text{P}_\text{n}=64\text{D}$
Let his near points be $x_n$, then
$\frac{1}{\text{x}_\text{n}}+\frac{1}{0.02}=64$
$\Rightarrow\ \frac{1}{\text{x}_\text{n}}+50=64$
$\Rightarrow\ \frac{1}{\text{x}_\text{n}}=14,$
$54=\frac{1}{\text{x}'_\text{n}}+\frac{1}{0.02}$
$\Rightarrow\ 54=\frac{1}{\text{x}'_\text{n}}+50$
$\Rightarrow\ \frac{1}{\text{x}'_\text{n}}=4$
$\Rightarrow\ \text{x}'_\text{n}=\frac{1}{4}=0.25\text{m}$
$\frac{1}{\text{f}}=\frac{1}{\text{f}_1}+\frac{1}{\text{f}_2}$
Also the effective power of the combination is given by
$P = P_1 + P_2$
- Let the power at the far point be $P_f$ for the normal relaxed eye of an average person.
Substituting $f_1 = 0.1$ and $\text{f}_2=\frac{1}{40}=0.02$, we get
$\text{P}_\text{f}=\frac{1}{\text{f}}=\frac{1}{0.1}+\frac{1}{0.02}=60\text{D}$
By the corrective lens the object distance at the far point is infinity.
The power required is given by
$\text{P}_\text{f}=\frac{1}{\text{f}'}=\frac{1}{\infty}+\frac{1}{0.02}=50\text{D}$
For eye + lens system, we have the sum of the eye and that of the lgasses $P_g$
$\therefore\ \text{P}_\text{f}=\text{P}_\text{f}+\text{P}_\text{g}$
$\Rightarrow\ \text{P}_\text{g}=\text{P}_\text{f}-\text{P}_\text{f}$
Substituting $P_f = 60D$ and $P'_f = 50D$, we get
$\text{P}_\text{g}=-10\text{D}$
- We are given that the power of accommodation of adult is 4D for the normal eye.
$\therefore\ 4=\text{P}_\text{n}-\text{P}_\text{f}$
or $\text{P}_\text{n}=64\text{D}$
Let his near points be $x_n$, then
$\frac{1}{\text{x}_\text{n}}+\frac{1}{0.02}=64$
$\Rightarrow\ \frac{1}{\text{x}_\text{n}}+50=64$
$\Rightarrow\ \frac{1}{\text{x}_\text{n}}=14,$
- Now, $P'_n = P'_f + 4 = 54$
$54=\frac{1}{\text{x}'_\text{n}}+\frac{1}{0.02}$
$\Rightarrow\ 54=\frac{1}{\text{x}'_\text{n}}+50$
$\Rightarrow\ \frac{1}{\text{x}'_\text{n}}=4$
$\Rightarrow\ \text{x}'_\text{n}=\frac{1}{4}=0.25\text{m}$
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