A patient can\'t see objects closer than 34.1 cm and wishes to clearly see objec

Question

A patient can't see objects closer than 34.1 cm and wishes to clearly see objects that are 22 cm from his eye.

(a) Is the patient nearsighted or farsighted?

(b) If the eye-lens distance is 2.09 cm, what is the minimum object distance p from the lens?
cm

(c) What image position with respect to the lens will allow the patient to see the object?
cm
(d) Is the image real or virtual? Is the image distance q positive or negative?

real image, distance is positive, distance is negative, virtual image

(e) Calculate the required focal length.
cm

(f) Find the power of the lens in diopters.
diopters

(g) If a contact lens is to be prescribed instead, find p, q, and f, and the power of the lens.

Explanation / Answer

a)

The person is farsighted , since he was able to see distant objects but unable to focus on objects at the normal near point for a human eye.

b)

With the corrective lens 2.09 cm in front of the eye, the object distance for an object 22 cm in front of the eye is

p=22-2.09 = 19.91 cm

c)

The upright, virtual image formed by the corrective lens will serve as the object for the eye, and this object must be 34.1 cm in front of the eye. With the lens 2.09 cm in front of the eye, the magnitude of the image distance for the lens will be

|q|=34.1-2.09 = 32.01 cm

d)

The image must be located in front of the corrective lens, so it is a virtual image and the image distance is negative .Therefore

q=-32.01 cm

e)

From thin lens equation

1/f =1/p +1/q

=>1/f =1/19.91 -1/32.01

f=52.67 cm

f)

The power of the corrective lens is

P=1/f =1/0.5267 =+1.9 Diopters

g)

With a contact lens, the lens to eye distance would be zero, so

P=22 cm ,q=34.1 cm

From thin lens equation

1/f =1/p +1/q

=>1/f =1/22 -1/34.1

f=62 cm

Power

P=1/f =+1.6 Diopters

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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