A hypermetropic person having near point at a distance of 0.75m puts on a spectacle of power 2.5D. The near point is now?
Answers
Hypermetropia is a condition of the eye where the person is not able to see things clearly when nearer to the eye. The normal near point of the eye is 25 cm , i.e. for normal individuals to see clearly, an object must be at a distance of atleast 25 cm. For a hypermetropic eye with a near point of 1m (100 cm), to see an object placed at 25 cm, the virtual image needs to be for med at 100 cm. Hence applying the formula,
1/focal length (f)= 1/ object distance (u) + 1/ image distance (v)
Since the image formed is virtual a - (minus) sign is assigned
Hence it becomes,
1/f = 1/u -1/v
which in this situation is,
1/f = 1/25 -1/100
1/f = 4/100 - 1/100 = 3/100
Hence f = 100/3 = 33.3 cm
Diopteric power of the eye P = 100 / f = 100/ 33.3= 3 D
Hence the answer is 3D
Answer:Hypermetropia is a condition of the eye where the person is not able to see things clearly when nearer to the eye. The normal near point of the eye is 25 cm , i.e. for normal individuals to see clearly, an object must be at a distance of atleast 25 cm. For a hypermetropic eye with a near point of 1m (100 cm), to see an object placed at 25 cm, the virtual image needs to be for med at 100 cm. Hence applying the formula,
1/focal length (f)= 1/ object distance (u) + 1/ image distance (v)
Since the image formed is virtual a - (minus) sign is assigned
Hence it becomes,
1/f = 1/u -1/v
which in this situation is,
1/f = 1/25 -1/100
1/f = 4/100 - 1/100 = 3/100
Hence f = 100/3 = 33.3 cm
Diopteric power of the eye P = 100 / f = 100/ 33.3= 3 D
Hence the answer is 3D