A patient can\'t see objects closer than 35 cm and wishes to clearly see objects

Question

A patient can't see objects closer than 35 cm and wishes to clearly see objects that are 19.2 cm from his eye.

(a) Is the patient nearsighted or farsighted?

farsightednearsighted   

(b) If the eye-lens distance is 1.95 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, 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.

diopters

diopters

Explanation / Answer

a)

He can not see the object closer than 35cm so, he is far sighthness

b)

P =19.2cm -1.95cm =17.25cm

c)

q =35cm -1.95cm =33.05cm

d)

Here the image and object are on the same side and distance is negative so virtual

e)

From the equation

1/f =1/p+1/q

p =17.25cm

q=33.05cm

then 1/f =1/17.25+1/33.05 =0.0579+0.0302=0.0881

f =11.343cm

f)

Then power =100/f =8.816diaopter

g)

In the case of contact lens

1/f =1/19.2+1/(-35)

1/f =0.0520-0.02857=0.02343

f =42.68cm

Then the power is =100/42.68 =2.343diopter

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