A woman wears bifocal glasses with the lenses 2.0 cm in front of her eyes. The upper half of each lens has power-0.500 diopter and corrects her far vision so that she can focus clearly on distant when looking through that half. The lower half of each lens has power +2.00 diopters and corrects her near vision when she looks through that half What are the far point and near point of her eyes?

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21-04-2020
Physics

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A woman wears bifocal glasses with the lenses 2.0 cm in front of her eyes. The upper half of each lens has power-0.500 diopter and corrects her far vision so that she can focus clearly on distant when looking through that half. The lower half of each lens has power +2.00 diopters and corrects her near vision when she looks through that half What are the far point and near point of her eyes?

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HELPPPPPP 15 POINTS!!!

moya1316 moya1316 · Answer 1 · 2020-04-22T14:18:37+02:00

Answer:

q = -2 m and q = -0.5 m

Explanation:

For this exercise we must use the equation of the optical constructor

1 / f = 1 / p + 1 / q

where f is the focal length, p and q are the distance to the object and the image, respectively

Let's start with the far vision point, in this case the power of the lens is

P = -0.5D

power is defined as the inverse of the focal length in meter

f = 1 / D

f = -1 / 0.5

f = - 2m

the object for the far vision point is at infinity p = infinity

1 / f = 1 / p + i / q

1 / q = 1 / f - 1 / p

1 / q = -1/2 - 1 / ∞

q = -2 m

The sign indicates that the image is on the same side as the object

Now let's lock the near view point

D = +2.00 D

f = 1 / D

f = 0.5m

the near mink point is p = 25 cm = 0.25 m

1 / f = 1 / p + 1 / q

1 / q = 1 / f - 1 / p

1 / q = 1 / 0.5 - 1 / 0.25

1 / q = -2

q = -0.5 m

the sign indicates that the image is on the same side as the object in front of the lens

andromache andromache · Answer 2 · 2022-03-02T06:13:36+01:00

The far point and near point of her eyes is q = -2 m and q = -0.5 m

Optical construction equation:

According to the above equation

we know that

1 / f = 1 / p + 1 / q

Here f is the focal length,

p and q are the distance to the object and the image

Since the power of the lens with respect to the far vision is -0.5D

Now we know that

power should be defined as the inverse of the focal length in meters

So,

f = 1 / D

f = -1 / 0.5

f = - 2m

Now

the object for the far vision point is at infinity p = infinity

So,

1 / f = 1 / p + i / q

1 / q = 1 / f - 1 / p

1 / q = -1/2 - 1 / ∞

q = -2 m

Now the near viewpoint is

D = +2.00 D

f = 1 / D

f = 0.5m

Since the near mink point is p = 25 cm = 0.25 m

Now

1 / f = 1 / p + 1 / q

1 / q = 1 / f - 1 / p

1 / q = 1 / 0.5 - 1 / 0.25

1 / q = -2

q = -0.5 m

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