A person is suffering from Hypermetropia. The near point of the person is 150cm. Calculate the focal length and the power of the convex lens used in his spectacles. Assume that the near point of a normal person is 25cm.
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
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