How Do Prescription Glasses Affect Vision Clarity at Various Distances?

[IntellectIsStrength]

Two questions... first one is a lot simpler but I'm still not able to do it for some reason?

1. Alex is prescribed new glasses because he finds it difficult to read his favourite Physics book at a convenient distance of 25 cm. His optometrist recommends lenses of power +2.0 Diopters. What is the maximum distance at which Alex can distinctly see objects (without his glasses)?

2. Samson needs glasses of power -7.5 Diopters. Removing his glasses, he looks into a microscope where a slide of amibae has been inserted. The slide stands at a distance of 1.2 cm from the objective lens of focal lengh 1 cm. The focal length of the eyepiece is 15 cm. In order for Samson to clearly see the amibae, he needs to rotate the knob on the side of the microscope. This adjusts the distane between the two lenses by having the eyepiece move while the objective stays fixed. Determine the separation distance between the two lense which Samson starts to distincly see the amibae.

Can you please just offer some hints as to how to approach the problems?

For the 1st one, I used the thin lense equation but I didn't get the right answer.

Thanks

 

[HallsofIvy]

My hint is look in your textbook for formulas relating focal length and object distance.

 

[jtbell]

Also, in eyeglass problems, we normally use infinity for the image distance, because that allows the person to view the image with completely relaxed eyes (no eyestrain).

 

[OlderDan]

Two questions... first one is a lot simpler but I'm still not able to do it for some reason?

1. Alex is prescribed new glasses because he finds it difficult to read his favourite Physics book at a convenient distance of 25 cm. His optometrist recommends lenses of power +2.0 Diopters. What is the maximum distance at which Alex can distinctly see objects (without his glasses)?

2. Samson needs glasses of power -7.5 Diopters. Removing his glasses, he looks into a microscope where a slide of amibae has been inserted. The slide stands at a distance of 1.2 cm from the objective lens of focal lengh 1 cm. The focal length of the eyepiece is 15 cm. In order for Samson to clearly see the amibae, he needs to rotate the knob on the side of the microscope. This adjusts the distane between the two lenses by having the eyepiece move while the objective stays fixed. Determine the separation distance between the two lense which Samson starts to distincly see the amibae.

Can you please just offer some hints as to how to approach the problems?

For the 1st one, I used the thin lense equation but I didn't get the right answer.

Thanks

The hint to search your textbook is a good one because some assumptions need to be made. For the first problem, in reality a person who needs +2.0 diopter glasses for reading might be able to see fine at long distance. The problem Alex has is not the inability to see clearly at large distances. It is the inability to see nearby objects clearly. He is farsighted. I'm bit puzzled why the problem is asking for a maximum distance. It would make more sense to me if they were asking you to find the minimum distance.

I don't know if you have learned about "accomodation" (adjustable focal length) and if you are supposed to take it into consideration in these problems. You might want to take a look at this discussion and see if any of this looks familiar.

http://www.physicsclassroom.com/Class/refrn/U14L6c.html

 

[IntellectIsStrength]

Thanks for the replies guys.

I just found out that I made the most farcical mistake. For the first question (the Alex one), all I had to do was use the thin-lens equation and solve for di. But I used the wrong value for f... I used 1/2 instead of 1/0.5 ! Since the power is 2.0 D, all I had to do to get f was to get the reciprocal of the power (which is 0.5).
So I solved that... the answer is 50 cm.

Still though, for number 2, I'm not quite sure what to do. Here's my attempt:

Givens:
P= -7.5 D
do= 1.2 cm = 0.012 m
f1= 1 cm = 0.01 m
f2 = 15 cm = 0.15 m

Firstly, I used the thin-lens equation to solve for di of the objective lens.

1/f = 1/do + 1/di
1/0.01 m = 1/0.012 m + 1/di
16.7 m = 1/di
di = 0.06 m

Now, here's the part I'm not sure about. I used the di for the objective lens (0.06 m) as the do for the eyepiece in order to solve for the di of the eyepiece. Is this correct?

1/f = 1/do + 1/di
1/0.15 m = 1/0.06 + 1/di
-10 m = 1/di
di = -0.1 m

I'm not quite sure what to do next... not even certain if what I did was right. Any help would be greatly appreciated.

Thanks in advance.

 

[OlderDan]

Thanks for the replies guys.

I just found out that I made the most farcical mistake. For the first question (the Alex one), all I had to do was use the thin-lens equation and solve for di. But I used the wrong value for f... I used 1/2 instead of 1/0.5 ! Since the power is 2.0 D, all I had to do to get f was to get the reciprocal of the power (which is 0.5).
So I solved that... the answer is 50 cm.

I'm still curious what you do with that focal length for the correcive lens to answer the question stated by the problem. What does it have to do with how Alex sees without his glasses?