MYOPIA (NEARSIGHTED)
(1) Person is able to see nearby objects clearly but is not able to see farway object
(2) Image is formed in front of retina
(3) It is corrected by using concave lens
$$\bbox[yellow, 5px, border: 2px solid red] {\dfrac{1}{f} = \dfrac{1}{\infty} + \dfrac{1}{-\text{PR}}}$$
Note:
\(\text{PR(Punctum Remotum)} = \text{person’s far point}\)
\(\dfrac{1}{\infty} = 0\)
EXERCISE
Q1
A person needs a lens of power −0.25 D for correction of his vision. What is the focal length of the corrective lens?
(A) −300 cm
(B) −400 cm
(C) −500 cm
(D) −600 cm
Q2
A patient with nearsightedness can’t see an object more than 200 cm clearly. The power of lens that can be used to help the patient to see clearly is…
(A) \(-0,5 \text{ D}\)
(B) \(-0,25 \text{ D}\)
(C) \(0,5 \text{ D}\)
(C) \(1 \text{ D}\)
Q3
A myopic person sees that her contact lens prescription is \(-1.25 \text{ D}\). What is her far point?
(A) 25 cm
(B) 40 cm
(C) 50 cm
(D) 80 cm
Q4
Joko has difficult seeing objects that are more than 2 meters away although he wears −2 Diopter spectacles. Calculate the power of the lens of his spectacles in order to see normally?
(A) −1 D
(B) −1.5 D
(C) −2 D
(D) −2.5 D
HYPEROPIA (FARSIGHTED)
(1) Person is able to see farway objects but are not able to see nearby objects
(2) Image is formed behind the retina
(3) It is corrected by using convex lens
$$\bbox[yellow, 5px, border: 2px solid red] {\dfrac{1}{f} = \dfrac{1}{25} + \dfrac{1}{-\text{PP}}}$$
Note:
\(\text{PP(Punctum Proximum)} = \text{person’s near point}\)
EXERCISE
Q1
A far-sighted person has his near point 40 cm, find the power of lens he should use to see at 25 cm, clearly.
(A) 0,5 D
(B) 1,0 D
(C) 1,5 D
(D) 2,0 D
Q2
The power of lens, a far-sighted person uses is 2.5 D. The minimum distance of an object which he can see without spectacles is…
(A) 45,80 cm
(B) 66,67 cm
(C) 79,87 cm
(D) 100,5 cm