1.If a nearsighted person has a far point df that is 3.50m from the eye, what is the focal length f1 of the contact lenses that the person would need to see an object at infinity clearly?

2.If a farsighted person has a near point that is 0.600m from the eye, what is the focal length f2 of the contact lenses that the person would need to be able to read a book held at 0.350m from the person's eyes?

Since the lens used for correcting near sighted person is concave lens therefore the image distance is negative due to virtual image produced by the lens.

We then have 1/f1 = 1/infinity -1/3.5

f1 = -3.5m

b) A far sighted person is a person that can only see far objects clearly but not near objects.

According to the lens formula

1/u+1/v = 1/f

Given u= 0.35m v = 0.6m f2= ?

1/f2 = 1/0.35 + 1/0.6

Since the lens used for correcting far sighted person is convex lens therefore the image distance is positive due to mostly real image produced by the lens (reason why the image distance is positive.

f2 = 0.22m

tifanihayyu tifanihayyu · Answer 2 · 2020-02-09T14:02:16+01:00

1. The focal length [tex]f_1[/tex] of the contact lenses that the person would need to see an object at infinity clearly is -3.5m

2. The focal length [tex]f_2[/tex] of the contact lenses that the person would need to be able to read a book held at 0.350m from the person's eyes is 0.22m

Explanation:

1.

A nearsighted person is the people that able to read a book or see a computer screen easily but they will have difficulty seeing things across the room. In other words, nearsightedness or myopiais a vision condition in which people can see close objects clearly, but objects farther away appear blurred.

Near point of distance from eye([tex]d_i[/tex]) = 0.60 m
Distance he can read ([tex]d_o[/tex])= 0.350 m

The focal length [tex]f_1[/tex] of the contact lenses that the person would need to see an object at infinity clearly is

[tex]\dfrac{1}{f} = \dfrac{1}{d_1} + \dfrac{1}{d_f}\\\dfrac{1}{f} = \dfrac{1}{infinity} + \dfrac{1}{-3.5}\\\dfrac{1}{f} = 0+ \dfrac{1}{-3.5}\\ f = -3.5[/tex]

2.

Farsightedness or hyperopia is a condition in which you can see distant objects clearly, but objects nearby may be blurry

u= 0.35m

v = 0.6m

[tex]f_2[/tex] = ?

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

[tex]\frac{1}{0.35} +\frac{1}{0.6} =\frac{1}{f_2}[/tex]

[tex]f_2 = 0.22 m[/tex]

The lens used to correct far sighted person is convex lens, therefore the image distance is positive because of real image mostly produced by the lens

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