Respuesta :
Answer:
(a). The eye's near point is 68.98 cm from the eye.
(b). The eye's far point is 33.33 cm from the eye.
Explanation:
Given that,
Power = 2.55 D
Object distance = 25 cm for near point
Object distance = ∞ for far point
Suppose where is the far point of an eye for which a contact lens with a power of -3.00 D is prescribed for distant vision?
(a) We need to calculate the focal length
Using formula of power
[tex]f =\dfrac{1}{P}[/tex]
Put the value into the formula
[tex]f=\dfrac{100}{2.55}[/tex]
[tex]f=39.21\ cm[/tex]
We need to calculate the image distance
Using formula of lens
[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}[/tex]
Put the value into the formula
[tex]\dfrac{1}{39.21}=\dfrac{1}{v}-\dfrac{1}{-25}[/tex]
[tex]\dfrac{1}{39.21}-\dfrac{1}{25}=\dfrac{1}{v}[/tex]
[tex]\dfrac{1}{v}=-\dfrac{1421}{98025}[/tex]
[tex]v=-68.98\ cm[/tex]
The eye's near point is 68.98 cm from the eye.
(b). We need to calculate the focal length
Using formula of power
[tex]f =\dfrac{1}{P}[/tex]
Put the value into the formula
[tex]f=-\dfrac{100}{3.00}[/tex]
[tex]f=-33.33\ cm[/tex]
We need to calculate the image distance
Using formula of lens
[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}[/tex]
Put the value into the formula
[tex]\dfrac{1}{-33.33}=\dfrac{1}{v}-\dfrac{1}{\infty}[/tex]
[tex]-\dfrac{1}{33.33}+\dfrac{1}{\infty}=\dfrac{1}{v}[/tex]
[tex]\dfrac{1}{v}=-\dfrac{1}{33.33}[/tex]
[tex]v=-33.33\ cm[/tex]
The eye's far point is 33.33 cm from the eye.
Hence, (a). The eye's near point is 68.98 cm from the eye.
(b). The eye's far point is 33.33 cm from the eye.
The near point for which a lens of +2.55 diopters is prescribed is 71 cm.
We have to obtain the focal length of the lens as follows;
f = 100/P
Where P is the power of the lens in diopters.
f = 100/2.55 = 39.2 cm
Now;
Given that;
1/f = -1/v + 1/u
f = focal length
v = image distance
u = object distance
1/v = 1/u - 1/f
u = Near point of the normal eye
1/v = 1/25 - 1/39.2
v = 71 cm
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