Question

A person with axial hypermetropia has a lens-retina distance of 1.9 cm and the maximum optical power of their eye is the same as that for a normal person.

A) what is the near point for this person?

B) What is the range accommodation this person needs to see objects from their near point all the way to their far point (Which is the same as for a normal eye)?

C) What is the optical power of the contact lenses used to treat this person and give them a normal near point of 25 cm>

Answer 1

Starting with hypermetropia which is a vision where the person can see clearly objects that are far away, but fail to focus objects near to their eyes. Humans with normal eyesight also posses a limit where they cannot focus if the object were closer. This is called the near point of the eyes. 25 cm is the near point for healthy, normal eyes.

Hypermetropia is a disorder that occurs when the image formed of the nearby object is behind the retina. This can be because of two reasons. Either the lens does not get strong enough or the lens to retina distance is smaller than normal. In our case, the hypermetropia is caused by shorter lens to retina distance, and not the power of the lens. The normal distance between the retina and the eye lens is 2 cm. For that case, the maximum power that a normal eye lens can achieve is when the object is at the near point and is in focus,

This is the lens equation, where

Which is the maximum power for a normal eye.

(A) For our hyperopic patient, near point will be further away, and to calculate it, we use

which is the near point for the person with hyperopia.

(B) The distance between the near point and the far point is called the range of accommodation. Hyperopic patients have the same far point as normal people and larger than normal near points.

At the far point, and for our guy,

So, the power at the far point is,

Hence, the powers can change from 52.63 to 54 Diopters for our patient which is the range of accommodation. In cm it is 73 cm to infinity.

(C) The solution to hypermetropia is a simple one. Use convex lens, such that they form a virtual image that acts as the object for the eye. For instance, in our case the near point is 73 cm. How will the person clearly see an object at 25 cm? Well, we use lens with a certain power such that the lens will take the object at 25 cm and form a virtual image at 73 cm. This will act as the object for the eye which will perceive the object to be at its near point and then see it clearly. Let us determine the power of the lens required for the person to decrease his near point to normal values,

There is a negative sign here because the image formed will be a virtual one, on the same side as the object.

A corrective convex lens of power 2.63D will give them a normal near point.