# Homework Help: Question about diverging lens for correcting myopia

1. Aug 28, 2013

### s3a

1. The problem statement, all variables and given/known data
The problem and its solution are attached as (the leftmost and middle) jpg files. (Given that the problem depends on the drawing, I think it's more convenient for the reader to view the text in the image as well.)

2. Relevant equations
N/A

3. The attempt at a solution
Editing the lens in the solution image, I obtain diverging_lens_made_by_editing.png. Basically, I am having trouble understanding why, for number 2, a lens that is convex on one side and concave on the other side works for solving myopia. Is it because, it is only expected for the light that the person is viewing to be diverged? Would the lens I made from editing work as well?

Any input would be greatly appreciated!

#### Attached Files:

File size:
16.3 KB
Views:
61
• ###### TheSolution.jpg
File size:
4.9 KB
Views:
125
File size:
931 bytes
Views:
136
2. Aug 28, 2013

### rude man

If you look at the convex-concave lens carefully you will notice that the concave radius is smaller than the convex one. Applying the lensmaker's formula will give you a net negative focal length, meaning a diverging lens.

3. Aug 29, 2013

### s3a

Is the radius the distance between the optical centre and the focal point?

Also, how can you tell the radius of the diverging lens part is larger than the radius of the converging lens part?

By the way, could the answer have been any of these three? (If so, why did they specifically choose the one they did?):
http://www.odec.ca/projects/2005/dong5a0/public_html/ConcvLens.png

Assuming that the radius of the converging lens part is smaller than that of the diverging lens part, then the role of the convex part is to decrease the amount that the light will be diverged by, right? Would that be why it was chosen out of those three from the link? That would explain why the convexo-convave lens was chosen over the double-concave one but, what about the plano-concave lens?

4. Aug 29, 2013

### rude man

I meant radius of curvature. The focal length depends on the radii of curvature and also the index of refraction.

Just by looking at it. And you have it backwards. The radius of the diverging (concave) lens is smaller than the other. That makes the net lens diverging. See http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lenmak.html

Make sure you get the signs of R1 and R2 right.

See the link above. Any lens is diverging if (1/R1 - 1/R2) < 0. (assumes n of glass > n of surroundings). So it can be plano-concave, concave-concave or convex-concave so long as the above inequality obtains.

5. Sep 2, 2013

### s3a

Okay so you're telling me to use $1/f = (n – 1) (1/R_1 – 1/R_2)$ but, what is $R = 2f$ for?

Could you please modify the image or something because, I still don't see it?

Do you think there is any logical reason for which the answer states only the convex-concave answer (despite all three of those being correct)?

6. Sep 2, 2013

### rude man

Where did R = 2f come from? Not from me.
Sorry, no. It's your image ... it's obvious to me that the concave radius is smaller than the convex. So, by the standard convention,
R1 = |R1|, R2 = - (-R2) = |R2|, but |R2| < |R1| so 1/R1 - 1/R2 < 0 meaning net divergence.

I don't see any. They're all diverging lenses. This is probably a manufacturability issue.