Physics optics question (Hypermetropia)?

The strength of a lens D is 1/f where f is the focal length in metres.

Use the lens formula 1/f = D = 1/u + 1/v

For a normal eye u is 0.25 m. For the bad eye u is 0.8 m. v i s 0.023 m in both cases.

For a normal eye this gives D as 47.48.

For the bad eye you get D as 44.73.

The strength of the correcting lens is therefore the difference between the two - i.e 2.75.

Ok, let's look at the myopic eye.

The eye can see clearly up to 25 cm. Note that this is the FURTHEST distance which can be seen clearly, not the least.

So, u = 0.25 and v = 0.023.

This gives D(myopic) = 1/0.25 + 1/ 0.023 = 47.48 dioptre.

A normal eye can focus at infinity.

So, D(normal) = 1/infinity + 1/0.023 = 43.48 dioptre,

The strength of the correcting lens is therefore D(normal) - D(myopic) = 43.48 - 47.48 = -4 dioptre.

Hope this helped. I notice that Depressonik has copied and pasted my answer! Maybe I should feel flattered.

I think it would be...

The strength of a lens D is 1/f where f is the focal length in metres.

Use the lens formula 1/f = D = 1/u + 1/v

For a normal eye u is 0.25 m. For the bad eye u is 0.8 m. v i s 0.023 m in both cases.

For a normal eye this gives D as 47.48.

For the bad eye you get D as 44.73.

The strength of the correcting lens is therefore the difference between the two - i.e 2.75.

for hypermetropia, if 'f' is the focal length of convex lens needed, then 1/f = 1/25 - 1/80

so that, f = 36.36 cm

and hence power of the lens is P = 1/f = 1/0.3636 = 2.75 D.

Near point is where the focus of the lens is and so

Focusing power of normal eye is 1/0.25(power = 1/focus in meter)

= +4D

Focusing power of the defective eye = 1/0.8

= +1.25D

For the defective eye to work properly and get focusing power of +4D it needs (+4D) - (+1.25D)

= +2.75D

Simple answer :)