Question

A myopic adult has a far point at 0.1 m. His power of accommodation is 4 diopters.

A.What power lenses are required to see distant objects?
B.What is the near point without glass?
C.What is his near point with glasses?

Take the image distance from the lens of the eye to the retina to be 2 cm.

Answer 1

A) Step 1: Find the power of the normal relaxed eye.

Given, far point for myopic adult, $0.1m$
This will be object distance, $u=-0.1m$
Image distance from the lens of the eye to the retina $=2cm=0.02m$ So, the required power$P=\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$P=\frac{1}{0.02}+\frac{1}{0.1}=60D$

Step 2: Find the power of the lens required to see the object at $\infty$

If the object is at ∞, then power required,

$P^{\prime }=\frac{1}{f^{\prime }}=\frac{1}{\infty }+\frac{1}{0.02}=50D$

This is the power of the eye+lens system, so the power of the lens

$P^{\prime }=P+P_g$

$50=60+P_g$

$\Rightarrow P_g=-10D$

Final answer: $-10D$

B) Step 1: Find the power for the far vision.

Given, far point for myopic adult $=0.1m$

This will be object distance, $u=-0.1m$

Image distance from the lens of the eye to the retina $=2cm=0.02m$

So, the power for the far vision,

$P_f=\frac{1}{f}=\frac{1}{v}\frac{-1}{u}$

$P_f=\frac{1}{0.02}+\frac{1}{0.1}=60D$

Step 2: Find the power for the near vision.

Given, power of accommodation $=4D$

Let the power of the normal eye for near vision be $P_n$.

Then

$4=P_n-P_f$

$\Rightarrow P_n=64D$

Step 3: Find the distance of near point without glass.

Let his near point be $x_n,$ then

$\frac{1}{x_n}+\frac{1}{0.02}=64$

$\frac{1}{x_n}+50=64$

$\therefore x_n=\frac{1}{14}\simeq 0.07m$

Final answer: $0.07m$

C) Step 1:Find the power at the far point for the normal relaxed eye.

Given, far point for myopic adult, $0.1m$
This will be object distance, $u=-0.1m$
Image distance from the lens of the eye to the retina $=2cm=0.02m$

So, the required power

$P_f=\frac{1}{f}=\frac{1}{v}\frac{-1}{u}$

$P_f=\frac{1}{0.02}+\frac{1}{0.1}=60D$

Step 2: Find the power of the lens+eye required to see the object at $\infty$

If the object is at ∞, then power required,

$P_f^{\prime }=\frac{1}{f^{\prime }}=\frac{1}{\infty }+\frac{1}{0.02}=50D$

Step 3: Find the power for the near vision.

Given, power of accommodation $=4D$

Let the power of the normal eye for near vision be

$P_n^{\prime }.$

Then $P_n^{\prime }=P_f^{\prime }+4$
$\Rightarrow P_n^{\prime }=54D$.

Step 4: Find the distance of near point with glasses.

Let his near point be $x_n^{\prime }$,
then $\frac{1}{x_n^{\prime }}+\frac{1}{0.02}=54$
$\frac{1}{x_n^{\prime }}+50=54$
$\therefore x_n=1/4=0.25m$

Final answer: $0.25m$