A person's prescription for her new bifocal glasses calls for a refractive power of -0.500 diopters in the distance-vision part, and a power of 1.75 diopters in the close-vision part. What are the near and far points of this person's uncorrected vision? Assume the glasses are 2.00 cm from the person's eyes, and that the person's near-point distance is 25.0 cm when wearing the glasses.

Question

contestada

Respuesta :

aristocles aristocles · Answer 1 · 2019-09-02T03:46:48+02:00

Answer:

Far point = 202 cm

Near point = 38.5 cm

Explanation:

For distance vision we know that

[tex]P = -0.500 diopter[/tex]

we know that

[tex]P = \frac{1}{F}[/tex]

[tex]F = \frac{1}{-0.500}[/tex]

[tex]F = -2 m[/tex]

now by lens formula we know that

[tex]\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{F}[/tex]

so we have an object positioned at infinite we have

[tex]\frac{1}{d_i} + 0 = \frac{1}{2}[/tex]

[tex]d_i = 200 cm[/tex]

now distance from eye is given as

[tex]d = 202 cm[/tex]

Now similarly for near point given at 25 cm from eye

the distance of the image should be 25 - 2 = 23 cm

power for near point is given as

[tex]P = \frac{1}{F}[/tex]

[tex]F = \frac{1}{1.75}[/tex]

[tex]F = 0.57 m[/tex]

now again by lens formula

[tex]\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{F}[/tex]

[tex]\frac{1}{d_i} + \frac{1}{23} = \frac{1}{57}[/tex]

[tex]d_i = 38.5 cm[/tex]