Question about diverging lens for correcting myopia

[s3a]

Homework Statement

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.)

Homework Equations

N/A

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!

 

[rude man]

1. 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!

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.

 

[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?

 

[rude man]

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

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

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

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.

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

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.

 

[s3a]

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

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

Just by looking at it.

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

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.

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)?

 

[rude man]

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

Where did R = 2f come from? Not from me.

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

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.

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)?

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