Lens Makers Equation Tutorial

[rbaker]

A video tutorial on the lens makers equation:

http://www.frostwire-preview.com/?ty...%3D-gu7BHAfvjo

[braheem24]

Thank You, My lab guys will be forced to watch this.

[wmcdonald]

There appears to be a virus associated with this file according to the folks at Norton. Proceed with care watching it.

[Craig]

click on the youtube and the virus goes away.

[tmorse]

A video tutorial on the lens makers equation:

http://www.frostwire-preview.com/?ty...%3D-gu7BHAfvjo

The Lensmakers Equation as written in this Ontario Georgian College video production uses measurement units in meters (m), rather than centimeters (cm) that this instructor prefers. The Lensmakers Equation used in this video…

1/l + 1/f = 1/l’

is valid only if the units of measurements used for l= Object distance and l’ = Image distance is meters (m), rather than centimeters (cm) that she prefers.


Vergence Formula:
L + F = L’

The object ray here has an incoming minus or negative vergence L in Diopters (D), strikes F, a +4.00lens, and then exits with plus or converging vergence L’ in D. to the right of the lens. The reciprocal of incoming L vergence is l or object distance, lens F reciprocal is focal length, and the reciprocal of exiting vergence of image L’ is image distance.
Since L = 1/l(m) F=1/f(m) and L’ = 1/l’(m)

L + F = L’ becomes

1/l(m) + 1/f(m) = 1/l’(m) or Lensmakers Equation used in this video.

If this opticianry instructor chooses to work in centimeters (cm) measurements rather than meters (m), her Lensmakers Equation must use 100 (the number of centimeters in a meter) rather than 1 (in meters) in the numerator of her Lensmakers Equation. Her Lensmakers Equation on the whiteboard should have read …


100/l(cm) +100/f(cm) = 100/l’(cm) or

100/-60cm + (100/25cm) = 100/l’cm

-1.67D + (+4.00D) = 100/l’cm


+2.33D = 100/l’cm


Thus Image distance l’cm = 100/+2.33D = +42.918cm to right of lens

[MakeOptics]

The Lensmakers Equation as written in this Ontario Georgian College video production uses measurement units in meters (m), rather than centimeters (cm) that this instructor prefers. The Lensmakers Equation used in this video…

1/l + 1/f = 1/l’

is valid only if the units of measurements used for l= Object distance and l’ = Image distance is meters (m), rather than centimeters (cm) that she prefers.

It works in cm, mm, m, dm, nm, um, etc. The unit of measure is not important in this equation as long as it remains consistent. When converting to power the unit does become important.

I would rather see the equation written with a little more accuracy showing it's roots to snell:

n/l + n/f = n'/l'

-or-

n/l + (n'-n)/r = n'/l'

It's a more useful formula in this sense, now we can measure a thin lens in water, or a thin lens with water at the front and air in the back. Just removing a slight assumption provides a more robust equation and relates back to it's roots.

If we look at snells law:

n * sin(i) = n' * sin(i')

if we want to develop a paraxial equation we can assume that any rays that are paraxial (parallel to the axis), must be close to the axis, so they are small and strike at an angle of incidence relatively low if we look at a table of angles (in radians) and compare them to their sine values the result would look like the graph below. What becomes apparent is that the angle in radians is negligable in difference to the actual sine of the angle (at low value like paraxial or close to the axis). So we can sub out the sin in snells law with the angle in radians

radian2ratio.xls
(added excel spreadsheet attachment, picture was hard to see)

n * i = n' * i'

The angle of incident is the angle the ray strikes in relation to the normal (perpendicular to the surface), if we were to draw a line through the radius that dissects the lens this line would be perpendicular to the lens surface and it's angle could be described as:

sin(a) = y/r

Since we have already made the assumption that the ratio and the radian measure of the angle can be subbed out interchangeably in the domain of y being small (paraxially), then:

n * (? -a) = n' * (? - a) = n * (y/l - y/r) = n * (y/l' - y/r)

If we multiple both side by the inverse height of the ray (again the assumption is that the ray entering the lens is leaving the lens at the same height, aka thin lens) then:

n * (1/l - 1/r) = n' * (1/l' - 1/r)

n/l - n/r = n'/l' - n'/r

moving things around so we can combine like terms

n/l + (n' - n)/r = n'/l'

Now we can understand the proof and see where assumptions are made and when the formula starts to degrade. In the day and age of paper and pen this equation was paramount to a lens designers success in the day and age of computers with 64 bit processors the need for assumptions and inaccuracy is ghost of a distant past, it's great for understanding but he underlying logic in the proof holds more value then the lensmakers equation. The lensmakers equation is a great tool for estimation, in this day and age and can be used on the fly (SALES FLOOR), but not as necessary in production (LAB).

[tmorse]

Nobody is arguing…. 42.91cm is the correct image distance. The issue is how this image distance was calculated, and to answer that we can break this question down in terms of vergence power.
All we need to picture is an object point located on the optical axis 60cm (or 0.60m) from lens, then strikes a +4.00 lens, and then emerges some distance from the back of this lens.
This video seems to suggests that is doesn’t matter what unit of measurement is simply plugged into Lensmakers Equation as written…
1/l + 1/f = 1/l’
But this formula is derived from Vergence Power Formula = L(in D) + F(in D) = L’(in D), which holds that an incoming vergence power L in Dipoters of vergence power (from object) strikes a (thin) lens F, and can exit the back of lens with only 3 possibilities… either a divergent (minus) ray, a parallel ray, or a converging (plus) ray, depending on the power of the incoming object vergence power (in D) and the F power (in D) it strikes.

Here, L + F = L’ and our object distance is 60cm = 0.60m. So here our incoming vergence L= incoming vergence in Diopters of an object 60cm (or 0.60m) away from lens 1/l(m) = 1/-0.60m = -1.66D, or 100/-60cm if we wish object distance to stay in cm = -1.66D
However, this Georgian College video appears to calculate this same incoming object vergence L in Diopters as
1/-60 or -0.0166D
Now our incoming vergence L (in Diopters) from object strikes our +4.00 lens power. L + F = L’ with L(incoming ray in D) + F (in D) = L’ (exiting vergence power L’ (in D) from back of lens.
L + F = L’ with object distance 60cm = 0.60m from the lens.

So incoming object vergence L (in Dipoters for object vergence power) is either:
a) Georgian College calculation for L=1/l = 1/-60cm or -0.0167D or,
b) BCCO calculation L= 1/l(m) = 100/l(cm) = 1/-0.60m or 100/-60cm= -1.67D
This incoming vergence power of object ray then strikes our +4.00 lens.
In Georgian College calculation…
L + F = L’
-0.0167D + (+4.00D lens) = L’(in D) = or +3.98D image vergence power which produces a real (plus) image a distance L’ = 1/l’ or 1’/+3.98D = +0.2512m or 25.13cm to right of this +4.00 lens.
In BC College of Optics calculation…
L + F = L’
1/-0.60m + (+4.00D lens) = L’(in D)
-1.67D +(+4.00) = L’ (in D) or +2.33D image vergence power which produces a real (plus) image at l’ = 1/L’ = 1/+2.33 = +0.4291m (or +42.19cm) to the right of this +4.00 lens, which is acknowledged as the correct image location l’.

Hopes this breakdown helps.

[MakeOptics]

Ugh, I see your point. Try to be a little more open minded and understand that rather than convert all measures in the equation in situ, she performed the conversion once when working with the diopter value. It's a valid mathematical method and if she used your updated version:

100/l + 100/f = 100/l'

If your version of the equation were multiplied by 1/100 o n both sides which is a perfectly valid mathematical operation and serves to simplify the equation then you end up with the same exact equation. If you insist that those components must be dioptric values which is what you are insisting by way of using vergence instead of focal lengths then the problem completely changes. Don't be so rigid in methodology it provides no additional value.

[tmorse]

Ugh, I see your point. Try to be a little more open minded and understand that rather than convert all measures in the equation in situ, she performed the conversion once when working with the diopter value. It's a valid mathematical method and if she used your updated version:

100/l + 100/f = 100/l'

If you insist that those components must be dioptric values which is what you are insisting by way of using vergence instead of focal lengths then the problem completely changes.

Sorry, there is nothing ‘updated’ here about the Lensmakers Equation formula:

a) 100/l + 100/f = 100/l’ or

b) 1000/l + 1000/f = 1000/l’ or

c) 1/l + 1/f = 1/l’ … the Lensmakers Equation used in this Georgian College video.

Only the unit of measurements for object distance and focal length of lens power F , is adjusted in this metric conversion… 1m = 100cm= I000mm. This conversion factor makes it easier for anyone having difficulty in making a metric conversions from cm to meters.

In a), if you use object distance in cm andlens focal length in cm, you can calculate image distance in cm.

In b) if you use object distance in mm and lens focal length in mm, you can calculate image distance in mm.
And in c), if both your object distance and focal length are measured in meters, your can calculated image distance l’ in meters (m) from back of lens.

There are no multiple ‘focal lengths’ in Lensmakers Equation… the only focal length there pertains to that of power F=1/f(m).
Neither l and l’ (object distance and image distance) have focal lengths, as they merely portray the relationship of vergence powers related to their object distance and to image distance respectfully through a certain media (here in air) .

Although 1/l and 1/l’ may appear to resemble a focal length like that in F= 1/f(m), both object distance l and image distance l’ are variable distances, and depend on the object distance l to produce an image distance l’ after going through F, the lens power involved. L describes the object’s ‘vergence power’ at a certain distance in front of this lens, and is given by object vergence power L=1/l(m) or (reciprocal) object distance l=1/L. And L’=1/l’ describes exiting vergence power of the image, used to determine image distance after this object ray had traveled through the lens. Incoming vergence L of an object 60cm from a lens can never equal 1/60, and F = 1/25 can never equal the focal length of a +4.00D lens. That is why it makes sense to solve this problem using vergence power formula... L + F = L' . The exiting vergence L' can then be converted to image distance l'=1/L'. Much easier in my mind.

The whole basis of modern optics theory is explained through this use of Vergence Power Theory through a simple thin lens. It becomes even more complicated when the ray is traced through a thick lens, with its lens center thickness and index of refraction.

Our BCCO students learn to ray trace through a thick lens, with given object distance and given object height. But this involves calculation and application of the lens’ principal planes and their nodal points. Vergence Theory and its application to thick lenses is clearly one of the trickiest part of our opticianry curriculum.

Does this help?