Question

A patient can't see objects closer than 32.0 cm and wishes to clearly see objects that are 20.0 cm from his eye.

(a) Is the patient nearsighted or farsighted?

nearsighted farsighted    

(b) If the eye–lens distance is 1.96 cm, what is the minimum object distance p from the lens? (Give your answer to at least three significant digits.)
cm

(c) What image position with respect to the lens will allow the patient to see the object? (Give your answer to at least three significant digits.)
cm

(d) Is the image real or virtual? Is the image distance q positive or negative? (Select all that apply.)

real image virtual image distance is positive distance is negative

(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.

p	=	cm	(Give your answer to at least three significant digits.)
q	=	cm	(Give your answer to at least three significant digits.)
f	=	cm
P	=	diopters

Answer 1

(a). The person is suffering from farsightedness.

(b). From the lens, the minimum distance of the object should be -(20-1.96)=-18.4cm.

(c). The desired image position from the lens is -(32-1.96)=-30.4cm.

(d). The image is virtual. Sign of p,q f depends on the convention. In my convention, I put objects to the left of the lens, and write all distances to the left as negative and all distances to the right as positive. So q will come out to be negative.

(e).

p=-18.4cm,

q=-30.4cm.

(f).

(g). p=-18.4cm,

q=-30m,

f=46.613cm,

P=0.021D.