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?

f1 = 3. 5m, f2= 0.84m

Explanation:

A nearsighted person will use a concave lens.

Using the formula:

1/o + 1/i = -1/f1(1)

o = object distance = 3.50 m

i = image distance = infinity

f1 = focal length

Substituting into 1

1/3.50 + 1/infinity = 1/f1

1/3.50 + 0 = 1/f1

f = 3. 50m

The negative means image is formed on the same side of the lens

Farsightedness is corrected with convex lens

So using

1/o + 1/i = 1/f2

i = image distance = -0.60m

The near point is taken as the image distance

o= 0.350m

The distance of the to the eye is considered the object distance.

1/f2 = 1/.350 - 1/0.60

1/f2 = 2.857- 1.6667

1/f2 = 1.190

f2 = 1/1.190

f2 = 0.840 m

the oxygenated blood then leaves the lungs through pulmonary veins, which return it to the left heart, completing the pulmonary cycle. this blood then enters the left atrium, which pumps it through the mitral valve into the left ventricle. from the left ventricle, the blood passes through the aortic valve to the aorta.

deoxygenated blood leaves through the right ventricle through the pulmonary artery. from the right atrium, the blood is pumped through the tricuspid valve (or right atrioventricular valve), into the right ventricle.