Question

How to determine image distance and object distance for hypermetropic eye

Answer 1

position the lens and white screen on the optical bench and place them so the distance fron the lighted object to the lens can be measured on the bench scale adjust the sreen to get a clear image

Answer 2

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