Glasses to fix near points and far points

  • Thread starter Thread starter vu10758
  • Start date Start date
  • Tags Tags
    Points
Click For Summary

Homework Help Overview

The discussion revolves around a problem involving optics, specifically the use of corrective glasses to adjust a person's near point and far point. The original poster presents two parts: calculating the power of glasses needed for a person with specific vision limitations and determining the new near and far points after using those glasses.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to apply the lens formula and power equations to find the necessary glasses for vision correction. They express confusion regarding the calculations for the new near point and far point after applying the glasses.

Discussion Status

Participants are exploring various aspects of the problem, including the setup of the lens formula and the implications of object and image distances. Some participants have offered insights into the calculations, while others are questioning the assumptions made regarding distances and the relationships between the variables.

Contextual Notes

There is mention of specific distances related to the person's vision, including their near point and far point, as well as the distance of the glasses from the eyes. Participants are also considering the implications of these distances on the calculations and results.

vu10758
Messages
95
Reaction score
0

Homework Statement



a)A person has a near point of .6m and far point of 2m . What power of glasses would be needed for the person to see infinitely far away. Assume that glasses are 2cm from the person's eyes. If the person used those glasses, what would his new near point be?

b) What power of glasses would be needed to bring the person's near point to .25m. If the person used those glasses, what would their new far point be?



Homework Equations



1/f = 1/o + 1/i
P=1/f

The Attempt at a Solution



This is what I have so far.

I know that
1/f = 1/o + 1/i

I want an image at infinity so 1/f = 1/o
The object is 2m from the eyes so it is 1.98 m from the glasses. The direction of the light is toward the glasses while the image is in the opposite direction. so o is negative.

1/-1.98 = -.505


And this is the correct answer according to the answer key.

I have problem solving for the new near point though.

1/f = 1/o + 1/i
-.505 = 1/o + 1/i

I want to create a new image for a distance of .6 m from the eyes or .58 from the glasses

-.505 = 1/.58 + 1/-i
-1/1.98 - 1/.58 = -1/i
I get i = .44 meters

but the correct answer is .82 meters. What did i do wrong?

I am having a similar problem for part b so I think I can get it if I find out what I did wrong here in part a.
 
Last edited:
Physics news on Phys.org
a) The object is at infinity (what the person wants to look at). The image needs to form at -1.98 meters (his far point, in front of the glasses).
 
Thanks for your help.

I just realized that I need help for part b as well.

b)

1/f = 1/o + 1/i

The person's far point is .6 or .58 from the glasses. I want him to see at .25 m or .23 m from the glasses.

1/-f = 1/.58 + 1/-.23
1/-f = -2.62
1/f = 2.62

that is the power. I am getting the wrong answer when I have to calculate the new near point though.

-2.62 = 1/1.98 + 1/-i

I get i = .32 m but the correct answer is .34m. Is there anything wrong with the 1/f and 1/1.98? I know that i/f =2.62 according to the answer key so I think I either rounded something wrong or have 1/1.98 wrong. However, I don't know what the object distance would be if it is not 1.98.

Thanks in advance for your help.
 
If -2.62 is 1/f and the answer is indeed .34 for as in the answer key then o must be greater than 3 meters away. How is that possible? Do I somehow combine that 1.98 m with the other part a, which is .82 meters?
 
a) For the new near point you have that 1/f = - 0.51 m
The image needs to form at a distance of 0.58 m in front of the lens in order for him to see it. That is i = - 0.58 m. The object distance that you get will then be his new near point. That is it is the new closest point (+ 2 cm) that he can hold something in front of him and still see it in focus with his glasses on.
 
Last edited:

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
5K
Replies
5
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K