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According to Pliny
01-21-2005, 08:07 PM
If I have a lens prescription of -5 diopters, how would that be expressed in terms of magnification? That is, a magnifying glass is usually sold as 2x power,5x power, etc.

barbitu8
01-21-2005, 08:59 PM
Diopter is the reciprocal of a focal lens, measured in meters. Prescriptive lenses are measured in diopters, but magnifying lenses (such as in binoculars) are measured in powers (magnified 2X, 3X, etc.)

For a spherical lens, all parts of the lens have the same power, whereas for cylindrical lenses, the power varies uniformly from zero in one meridian to the dioptric value in the perpendicular meridian.

The third number is the axis or orientation of the cylindrical component.

SavageNarce
01-21-2005, 09:30 PM
From this site: (http://www.bertech.com/product8/magnifi_notes.htm)
Diopter is a measurement of a lens’ ability to bend the light of a viewed object and thereby increase its apparent size. Each diopter represents a 25% increase in the size an object is magnified. A four diopter lens (referred to as a 4d lens) would increase the size of an object by 100%. Viewing a 2” wide object under a 3 diopter (75% increase) magnifier, would increase the size of the object by 1.5” and cause the object to appear to the 3.5” wide. The diopter of a lens can be calculated by dividing the number 40 by the focal length (in inches) of a lens. Once the diopter number is known, other parameters such as magnification power and total power can be calculated.

Magnification power is a measure of how much larger an object is after magnification. A magnification power of 1x means the object’s size has been increased by 100%. Our 2” object seen through a lens with a magnification power of 1.5x would appear to have added 150% on to its original size. So the 2” object will appear through the lens to have increased in size 3”. Its new size would be 5 inches (2” original size + 3” increase in size). Since the magnification power represents 4 times the diopter number, divide diopter by 3 to calculate it.

Q.E.D.
01-22-2005, 12:43 AM
Since it hasn't been mentioned yet, I'll discuss the difference between negative and positive diopter numbers. A positive number designates a convex lens, also known as converging lens - since the light rays passing through it converge towards the center axis. These lenses magnify objects seen through them. For eyeglasses, converging lenses are used to correct farsightedness.

A negative diopter, on the other hand, specifies a concave, or divergent lens. Divergent lenses spread the light rays apart, and make objects seen through them look smaller. They are used in eyeglasses to correct nearsightedness

According to Pliny
01-22-2005, 08:52 AM
...Since the magnification power represents 4 times the diopter number, divide diopter by 3 to calculate it.

Now that doesn't make sense, does it?

So what's my answer?
My prescription is -5 diopters, so what is its magnification? Would it be -5/4x or -5/3x?

Happy Fun Ball
01-22-2005, 09:46 AM
Actually the magnification of a lens is always 1, irregardless of the focal length or diopters of the lens. Before you answer what the magnification is, you need to specify the rest of the system. For example, if the object you are imaging is 10 meters away, to satisfy the imaging condition the lensmaker formula must be satisfied:

1/t1+1/t2=1/f=D=-5

Where t1 is 10 meters. This make the image plane -10/51 meters behind the lens (this is a virtual image). The magnification of the system is given by t2/t1, or in this case -1/51. Try this again with the object at 20 meters and you get a magnification of -1/101.

When you are talking about eye glasses, it gets more complicated because there is a lens in the eye (with a varying focal length) that must also be taken into account in your system.

According to Pliny
01-22-2005, 10:26 AM
Sorry, I don't buy that.
A lens in a pair of glasses has a certain curvature that would be the same as some magnifying glass.
This is true no matter whether the glasses are worn or not.
It's magnification has nothing to do with the lens of the wearer.

Happy Fun Ball
01-22-2005, 12:29 PM
Sorry, I don't buy that.
A lens in a pair of glasses has a certain curvature that would be the same as some magnifying glass.
This is true no matter whether the glasses are worn or not.
It's magnification has nothing to do with the lens of the wearer.
Sorry, but it is true...

Probably what you are asking about is the visual magnification. The visual magnification is defined as:

1+Dnp/Flens

Where Dnp is the near point of the eye (the closest that someone can focus, typically about 0.2 m) and F is the focal length of the lens in question. This equation is technically only valid if the lens is very close to the eye. For your lens, the focal length is -1/5.

To show you that the magnification of a magnifying glass is related to the eye, try looking at different objects with the lens at different distances from the eye (the distance from the lens to the object you are looking at will also change to fulfill the lensmaker formula above). You will notice that the magnification changes (i.e. the image gets bigger and smaller) depending on the distances between the object lens and eye.

Binoculars and Telescopes are typically afocal arrangements (4-f systems) that have an angular magnification defined by -f1/f2 (This is simplistic, the eye relaxes at focal lengths of 1 m, so telescopes are usually built to accomidate this. Also I am talking about Keplerian configurations. If I didn't include this I am sure someone would be along to pick my nit.)

I am pressed for time (walking out the door), but I'll try to come back and explain better this afternoon.

barbitu8
01-22-2005, 02:16 PM
Sorry, I don't buy that.
A lens in a pair of glasses has a certain curvature that would be the same as some magnifying glass.
This is true no matter whether the glasses are worn or not.
It's magnification has nothing to do with the lens of the wearer.
Prescriptive lenses have three variables: sphere, cylinder, and axis. A magnifying glass has only one variable.

According to Pliny
01-22-2005, 02:37 PM
Sorry, but my lenses are spherical, there is no cylinder or axis.
They are plain lenses, not aspherical. No need to try to complicate this question.
It should be a simple calculation. Such as the divide by 3 (or is it 4) mentioned by SavageNarce above.

And my eyes or anyone elses eyes are not part of the question or answer.

Think of the lenses as sitting on a test bench in a lab, aiming a light beam at a target or something.

A magnifying lens of 2x will give you a certain pattern on the target.

A prescrition lens can be ground to give the identical effedt.

What is the prescription that matches 2x?

And what magnification matches -5 diopters?

barbitu8
01-22-2005, 02:49 PM
Diopters are based on the focal point of the lens, and the magnification will vary according to how far the lens is from the object. Magnification lenses are afocal. See Eyer8's prior post.

Happy Fun Ball
01-22-2005, 06:15 PM
And my eyes or anyone elses eyes are not part of the question or answer.

Think of the lenses as sitting on a test bench in a lab, aiming a light beam at a target or something.

A magnifying lens of 2x will give you a certain pattern on the target.

A prescrition lens can be ground to give the identical effedt.

What is the prescription that matches 2x?

And what magnification matches -5 diopters?

Sorry, but it is really part of the question and answer (trust me, I spent a lot of money to be educated on this topic)...

Sorry for the rambling nature of this post…

A lens with a 2 x magnification is meaningless; as I said earlier you need to know the distances to be able calculate the magnification.

To take a picture with a camera you need to fulfill the imaging condition, meaning the inverse distance from the lens to the object you are photographing (say a flower) plus the inverse distance from the lens to the film must equal the power of the lens (i.e. inverse focal length) in diopters. This is the lensmaker (or thin lens) equation:

1/t1 + 1/t2 = 1/f

The magnification of the system is defined as t2 (the distance from the lens to the film) divided by t1 (the distance to the flower from the lens). If this equation is not fulfilled, you cannot take a picture, it will be out of focus and blurry. Thus you cannot define the magnification of the lens without knowing where object plane and image plane are.

(Nitpick: Of course this is a simplistic view (but probably good enough). I am treating lenses as infinitely thin and aberration free. They are not.)

You can play with magnification at this site (http://www.matter.org.uk/tem/lenses/thin_lens_optics.htm).

You can also find out lots by going to this really good site (http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lenscon.html#c1).

With the eye it is more complicated. With the eye you already have a variable focal length lens about 2 cm from the retina (the film). This lens must be taken into account when analyzing the optical system. A standard analysis has the lenses from the glasses (or what have you) placed directly against the eye (lenses right next to each other can be combined by noting that their powers (in diopters) add, thus you can treat them as a single lens). When you do this, you can find a magnification given by the equation above (the “visual magnification” of a lens):

Mv = 1 + Dnp / f (object at near point)

This result is obtained by using an object at the nearest point you can focus on with the glasses (i.e. Dnp has shifted due to the extra lens). When you carry this analysis out using objects at infinity (say mountains in the far distance, anything above a couple hundred feet is infinity for us optics guys typically), you will find the magnification is:

Mv = Dnp/f (object at infinity)

So there you have two limits of your magnification (0 and -1) for your -5 dipoter lens and a typical near point distance of 20 cm.

The focal length of the eye lens can be limited by noting that when looking at an object at infinity, the focal length must be ~2cm (the distance from the lens to the retina – plug the values into the lensmaker equation). The other limit is the near point distance I mentioned above. This is the closest a person can focus, typically about 20 cm. To find the focal length in this case, plug Dnp and the 2cm distance from lens to retina into the lensmakers equation and solve for f. You can also use the focal length for a object distance of ~ 1m. This is the object distance for most people when the eye is relaxed. To look it objects closer or further you will need to work your eye.

Hope this helps. Either way I am off again…

According to Pliny
01-22-2005, 07:15 PM
Thank you SavageNarce !
Upon rereading the answer I realize why dividing by 3 works.
So the answer is simple: Divide diopters by 3 to get magnification.
[Eyer8 - I have the answer to my answer, and if you want to continue to deny the connection between the two units, go right ahead, but you'll exuse me if I won't check back to see how your own arguments on your own questions are progressing. ]

Q.E.D.
01-22-2005, 07:54 PM
Eyer8 is correct. However, it's important to note that for magnifying glasses, the image plane is considered to be the typical minimum focus distance of the eye, which is normally taken to be 250 mm. Once you know that, you can reduce his equations to I = 250 / F, where I is the image magnification and F is the lens focal length (in mm), for an eye focused to infinity. This is the same way magnifying glass manufacturers specify power. To work this out in dopters, use I = .250 x D - which is the same as dividing diopter by 4.

Q.E.D.
01-22-2005, 08:07 PM
...for an eye focused to infinity.
Scratch that bit. I'm conflating two different concepts here. The above equation is, of course, based on an eye focused to 250 mm - the minimal focal distance referred to previously.

Happy Fun Ball
01-22-2005, 10:49 PM
Thank you SavageNarce !
Upon rereading the answer I realize why dividing by 3 works.
So the answer is simple: Divide diopters by 3 to get magnification.
[Eyer8 - I have the answer to my answer, and if you want to continue to deny the connection between the two units, go right ahead, but you'll exuse me if I won't check back to see how your own arguments on your own questions are progressing. ]
Hmmm...

Good luck [b]Pliny[/p], I was only trying to help.

hammerbach
01-22-2005, 11:33 PM
I've been wondering about this ever since I got a prescription I could read, so this thread had been very useful. But could somebody please tell me what "cylinder" denotes?

Q.E.D.
01-22-2005, 11:43 PM
I've been wondering about this ever since I got a prescription I could read, so this thread had been very useful. But could somebody please tell me what "cylinder" denotes?
Cylinder is the aspherical component needed to correct for astigmatism. It means the curvature along one axis is different than the curve along the other, and the number specifies this difference in diopters. The axis specification denotes the angle from the horizontal that the cylinder is oriented.

Happy Fun Ball
01-23-2005, 12:00 AM
I've been wondering about this ever since I got a prescription I could read, so this thread had been very useful. But could somebody please tell me what "cylinder" denotes?

Barbitu8 had it right in post 2. The focal length (or power) of a lens is dependent on the curvature of the two surfaces. You can visualize a lens as the intersection of 2 spheres have radii R1 and R2. The power of a lens (in diopters) is proportional to 1/R1 + 1/R2 (for a double convex lens only - you have to be careful of sign conventions here).

Now imagine that the lens is deformed and is the intersection of 2 ellipses or a sphere and an ellipse. In this case, the lens will have a different power along one axis of the ellipse (because of the different radius) than along the other. The cylinder is a measure of this difference in power while the axis number gives the orientation of the ellipse.

Happy Fun Ball
01-23-2005, 12:01 AM
Or what QED said...

hammerbach
01-23-2005, 09:56 AM
Thank you!

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