A patient can\'t see objects closer than 49.2 cm and wishes to clearly see objec
ID: 1509515 • Letter: A
Question
A patient can't see objects closer than 49.2 cm and wishes to clearly see objects that are 20.0 cm from his eye.
(a) Is the patient nearsighted or farsighted?
(b) If the eye–lens distance is 1.99 cm, what is the minimum object distance p from the lens? (Give your answer to at least three significant digits.)
(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.)
(d) Is the image real or virtual? Is the image distance q positive or negative? (Select all that apply.)
(e) Calculate the required focal length.
(f) Find the power of the lens in diopters.
(g) If a contact lens is to be prescribed instead, find p, q, and f, and the power of the lens. (Give your answer to at least three significant digits.)
Explanation / Answer
Here,
a) as the patient is not able to see the distant object
he is farsighted.
b)
object distance , do = 20 - 1.99
do = 18.01 cm
the minimum object distance is 18.01 cm
c)
for the lens
image distance ,di = - 49.2 + 1.99
di = -47.21 cm
the image distanc is -47.21 cm
d)
the image is virtual
e)
Using lens formula
1/f = 1/di + 1/do
1/f = 1/18.01 - 1/47.21
f = 29.11 cm
the focal length required is 29.11 cm
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