An eye has a near point distance of 0.75 m. What sort of lens in spectacles would be needed to reduce the near point distance to 0.25 m? Also calculate the power of lens required. Is this eye long-sighted or short � sighted?
Answers
Answer:
To reduce the near point distance to 0.25m the hypermeteropic person who has the near point of 0.75m will need a converging spectacles.
To calculate the power of the spectacles we have to determine the focal length at first.
u= 0.25m ( At normal near point distance of object).
v= 0.75m (Defective eye's near point in front of lens).
So according to the formula:
1/f= 1/v - 1/u
1/f= (1/-0.75) - (1/-0.25)
1/f= (-1+3)/0.75
f= 2.67
So focal length of the eye is 2.67.
According to the problem , the power required would be :
P = 1/f in meters
P= +0.37 D.
The eye is long sighted.
Explanation:
The Lens formula is given as:
The distance of the image = 0.75m
The distance of the object = 0.25m
f = 2.67
Then the power of the eye is the inverse of the focal length of the eye.
P = 0.37D