A nearsighted person has near and far points of 10.0 cm and 20.0 cm, respectivel

Question

A nearsighted person has near and far points of 10.0 cm and 20.0 cm, respectively, when she does not wear any corrective lenses. Her contact lenses have a power equal to 4.75 diopters, which is not strong enough to completely correct her vision.

(a) Calculate the farthest distance from her eyes that an object can be placed and still be in focus when she is wearing the contacts. Explain clearly. (Hint: the image of this object must be formed at 20.0 cm from her eyes.)

(b) Calculate the closest distance she can hold a book and still be able to read it, when she is wearing the contacts.

Explanation / Answer

Let's start by calculating the focal length of contact lenses

D = 1/f in meters

f= 1/D

f = 1/ 4.75

f = 0.2105 m

f = 21.05 cm

Part a)

We use the expression for thin lenses

1/f = 1/o + 1/i

1/o = 1/f – 1/i

The image must be at the greatest distance that the eye can focus

i = 20 cm

1/o = 1/21.05 - 1/ 20= (-20-21.05)/20* 21.05

1/o = -1.05/421

o = -400.95 cm

This is the greatest distance to the object can be placed and see nitidamente

Part b)

We use the same expression, but in this case the image must be the minimum clear distance from the eye, points of view

i = 10 cm

1/o = 1/f – 1/i

1/o = 1/ 21.05 – 1/10 = (10-21.05)/ 10*21.05

1/o = -11.05 /210.5

o = - 19.05 cm

I have a question for the correction of myopia divergent lenses are used so the power was being given by -4.75

I'll repeat the calculations with this I think is right.

. f = -21,05 cm

Part a)

1/o = -1/21.05-1/20 = - 41.05/(-421)

o = 10.26 cm

Part b)

1/o = -1/21.05-1/10 = -31.05/(-210.5)

o = 6.78 cm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.