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A myopic (nearsighted) person has a far point distance of 150 cm. What is the refractive power of his ideally prescribed eyeglasses if he wears his eyeglasses a distance of 3.0 cm from his eyes?

Respuesta :

Answer:0.68 D

Explanation:

Given

Person has a far point distance of 150 cm

i.e. [tex]d_i=-150 cm[/tex]

therefore

v=-150+3=-147 cm

From Lens Formula

[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}[/tex]

Here the object distance is infinity

[tex]\frac{1}{f}=\frac{-1}{147}+\frac{1}{\infty} [/tex]

Therefore f=-147 cm

and [tex]f=\frac{1}{P}[/tex]

[tex]P=-\frac{100}{147}[/tex]

P=-0.68 D

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