# Homework Help: Near-sighted lens fix

1. Jul 18, 2007

### GravityGirl

A 77 year old man has a near point of 115 cm. What power lens (in diopters) does the man need in order to be able to see an object clearly at the normal near point of 25 cm? Assume that the man will get contact lenses, that is neglect the two centimeters that glasses would be away from the eye.

ok so the goal here is to move the objects image from 115 cm to 25 cm.

i am going to use the equation 1/s+1/s'=1/f and 1/f=P

so this looks really simple, but i am not getting the right answer. for 1/f or P i am getting 4.87 d. and i converted to meters before i did the calculation. please help me.

2. Jul 18, 2007

### Staff: Mentor

I'm not familiar with the term near point (sorry), but are the numbers really 25cm and 115cm? Those seem around 10x too big. If the near point is the focul distance of the human eye, that's more like a 2.5cm number....

3. Jul 18, 2007

### GravityGirl

i tired 2.5 cm and 11.5 cm and that didn't work. any other suggestions

4. Jul 18, 2007

### Staff: Mentor

What's the definition of near point?

5. Jul 18, 2007

### GravityGirl

near point is the nearest point of an object that a relaxed eye can see. usually for a normal eye it is about 25 cm.

6. Jul 18, 2007

### G01

The average near point for a human is indeed 25cm.

You want the man to be able to read something 25cm away from his face like a normal person right? That way he won't have to hold a newspaper 115cm away from his face to read it!

We know the closest he can see is 115cm, and we want him to be able to see an object at 25cm. Now, the object is at 25cm, so its obvious what the object distance is. What would the image distance be if we want the image to appear at his near point?

7. Jul 19, 2007

### andrevdh

The object ditance will be 25 centimeters. That is he wants to hold the object at a distance of 25 centimeters from his eyes when he looks at it. The image needs to form at a distance of 115 centimeters from his eye. This means that the image distance will be - 115 centimeters (the contact lens forms a virtual image at this position which enables the senior citizen to see it with his unaided eyes). The function of the contact lens is therefore to shift the object from 25 to 115 centimeters for his eyes to be able see it.

You just need to convert these distances to meters before inserting them into the equation in order to get the answer in diopters.

Last edited: Jul 19, 2007