A person cannot see objects clearly when they are placed at a distance less than 30 cm from the what should be the power of corrective lens and what type of lens is used to correct the defect
Answers
Answered by
27
The near point of a normal eye is equal to 25 centimeters.
According to the question the person is not able to see the objects clearly when they are placed at a distance less than 30 cm from the eye.
Hence, the person is not able to see the objects placed at the near point of his or her eye.
Thus,
the person is suffering from the defect of vision called as hypermetropia.
HYPERMETROPIA :-
In this type of defect of vision the person is not able to see the nearby objects clearly but is able to see far objects clearly. In this type of defect of vision the near point of a person goes far.
CAUSES OF HYPERMETROPIA :-
☸️ The eye lens becomes thin that is the focal length of eye increases.
☸️The eyeball contracts.
CORRECTION OF HYPERMETROPIA :-
HYPERMETROPIA can be corrected using a convex lens of appropriate focal length.
POWER OF LENS :-
So,
The power of lens will be +3.33 diopter.
Hope this will be helping you!
Thanks!
According to the question the person is not able to see the objects clearly when they are placed at a distance less than 30 cm from the eye.
Hence, the person is not able to see the objects placed at the near point of his or her eye.
Thus,
the person is suffering from the defect of vision called as hypermetropia.
HYPERMETROPIA :-
In this type of defect of vision the person is not able to see the nearby objects clearly but is able to see far objects clearly. In this type of defect of vision the near point of a person goes far.
CAUSES OF HYPERMETROPIA :-
☸️ The eye lens becomes thin that is the focal length of eye increases.
☸️The eyeball contracts.
CORRECTION OF HYPERMETROPIA :-
HYPERMETROPIA can be corrected using a convex lens of appropriate focal length.
POWER OF LENS :-
So,
The power of lens will be +3.33 diopter.
Hope this will be helping you!
Thanks!
Attachments:
Answered by
2
AnswerAnswer power of lens will be 0.67D in positive
Explanation:
Similar questions