Seneca or Nait for dual license ?

[tmorse]

Thank goodness they're not offering flight school lessons.

Har Har!!
Hey, better learn your history. In WW2 aviators were taught to fly an airplane, understand weather patterns, learn deflection angles for shooting their machine guns, parachuting, and who knows what else… all in only 3-months. Sure it was wartime, but they were motivated, and their necks were on the line.

We are dealing here with opticianry, where many the optical staff in Optometrists office have only a few weeks of in-house training, and yet still manage. And many choose to take a 2-year program. Good for them. Hire them. The NAIT program takes 4-years, so what? Hire them instead. The new ‘Online’ opticianry programs can take as little as 55-weeks. So what? This is all process.
But in Canada we have rigorous National NACOR exams to weed out anyone unqualified to dispense. Years ago Ontario had a mandatory written exam for licensing that dealt with optical formulas. Our BCCO students are taught these various geometric Optics formulas to be able to troubleshoot, together with the required hands-on competencies… and in only 6-months. This is our private opticianry school’s proprietary process.

[MakeOptics]

Another optical program proof:

[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 need to 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, 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 vergences power and the F power 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 then 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 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.

[Lab Insight]

Har Har!!
Hey, better learn your history. In WW2 aviators were taught to fly an airplane, understand weather patterns, learn deflection angles for shooting their machine guns, parachuting, and who knows what else… all in only 3-months. Sure it was wartime, but they were motivated, and their necks were on the line.

We are dealing here with opticianry, where many the optical staff in Optometrists office have only a few weeks of in-house training, and yet still manage. And many choose to take a 2-year program. Good for them. Hire them. The NAIT program takes 4-years, so what? Hire them instead. The new ‘Online’ opticianry programs can take as little as 55-weeks. So what? This is all process.
But in Canada we have rigorous National NACOR exams to weed out anyone unqualified to dispense. Years ago Ontario had a mandatory written exam for licensing that dealt with optical formulas. Our BCCO students are taught these various geometric Optics formulas to be able to troubleshoot, together with the required hands-on competencies… and in only 6-months. This is our private opticianry school’s proprietary process.

Training in 1939 to 1945, although quick was meant to get planes in the air. while their missions were meant to drop bombs, not carry paying passengers to multiple destinations.

I've seen first hand over the years the results of organically grown opticians from BC, NAIT, Seneca and Georgian. I do have to say there's both good and bad from every training organization.

[tmorse]

I've seen first hand over the years the results of organically grown opticians from BC, NAIT, Seneca and Georgian. I do have to say there's both good and bad from every training organization.

I agree... 50% of all opticianry students fall into the bottom half of their class.

[Lab Insight]

I agree... 50% of all opticianry students fall into the bottom half of their class.

If I had to pick one that produces the best crop, Georgian would prevail (although they're desperately in need of a management change and major overhaul).

[MakeOptics]

Nobody is arguing…. 42.91cm is the correct image distance. The issue is how this image distance was calculated, and to answer that we need to 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, 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 vergences power and the F power 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 then 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 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.

You argued that her formula was inaccurate when using cm and even gave an alternative for that converted the formula to stick to the rigid implementation that it must be meters by placing a 100 in the numerator. She used ratios in her video using focal length in cm and even explained she prefers working in cm. Your post suggests that it must be converted to vergence, which is an additional discretization that you are interjection so that the formula doesn't work. That is why I mentioned in the post I linked to the derivation that bounds like domain, range, and yes unit of measure are important but she is correct in saying that as long as the units are the same it does not matter. In my linked post I outlined my personal preference where the index is a variable in the equation, which would also make her formula and yours wrong in certain situations.

My point being her math and understanding is correct, your conversion to vergence is unnecessary in this case and only creates more confusion. However I can throw away my math boops and calculators and concede that the only true way of computing is by using on unit of measure and vergence. Good luck with your program.

[MakeOptics]

The video never talks about vergence, the video never puts a D as a unit of measure, and never even converts the components to Diopter those steps are unnecessary since the question never asks for diopter as a response. The questions asks for distances which are expressed in mm, cm, dm, m, um, nm, inches, feet, yards, miles, etc. I can agree that convention dictates to use specific units but her understanding of the equation overcomes any pitfalls that may arise when working in non conventional units. If that is wrong writing 1/l as outlined in my other post is wrong as well since the real version should be n/l, which is just as wrong since the real equation should be n * (y/l - y/r) = n' * (y/l'), which is wrong since the real equation should read n * i = n' * i', which if you can guess is wrong since it should read n * sin(i) = n' * sin(i').

It all comes back to sells law and a fundamental building block for all of these formulas, every time we change that formula, to apply to guassian optics, to simplify the index to 1 since we are dealing with a thin lens in air, to remove the power of the lens thickness to simplify the math all of these bounds and assumptions should in my opinion be outlined. This formula is specific in that it only works withing 15 degrees at most of the optical axis until it loses accuracy. The instructors converted from diopter to focal length in cm and explained it as a preference, I prefer working in mm so that's how I work things and most the time I work in mm unless it's cumbersome to do so then I work in whatever unit suits me.

[MakeOptics]

You know the more I think about it, you are correct. The equation is wrong,it is a horribly outdated cheat to the sells formula. I have issue with many of these descritized versions of laws in physics that are distorted versions of truth.

I also prefer vergence, however I used swine step along method to ray trace and never use thin lens equations and never convert to diopters. I can remove thickness from focal lengths through a lens material, but I have to convert to diopters and use a clunky formula to do it in diopters.

[tmorse]

You know the more I think about it, you are correct. The equation is wrong,it is a horribly outdated cheat to the sells formula. I have issue with many of these descritized versions of laws in physics that are distorted versions of truth.

I also prefer vergence, however I used swine step along method to ray trace and never use thin lens equations and never convert to diopters. I can remove thickness from focal lengths through a lens material, but I have to convert to diopters and use a clunky formula to do it in diopters.

Glad you see my point.

And yes, we also teach BCCO students ray tracing through thick lenses using vergence, material index, and use object/image distances and lens thickness measurements in meters.

Image distance for a thick lens is fairly straightforward, but image height and magnification require more intricate formulas to calculate and use position of both lens’ principal planes and their nodal points in our calculations, and through both plus and minus (thick) lenses. This is why our BCCO students consider ‘vergence’ to be the trickiest portion of our geometric optics curriculum.

[MakeOptics]

Glad you see my point.

Just don't want to argue.

[MakeOptics]

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).

Nobody is arguing…. 42.91cm is the correct image distance. The issue is how this image distance was calculated, and to answer that we need to break this question down in terms of vergence power.

I must be misunderstanding.

1/l (cm) + 1/f (cm) = 1/l'(cm) L + F = L'

That's what you are saying which is correct, I am saying that the vergence was never discussed except to point out that using cm which was her preference does not work when converting to diopters. She already covered your argument (minute marker 1:12, 3:00, and 12:40),

[tmorse]

I will try one more time.
Perhaps I should not have said we “need” to use vergence, but rather that you “can” use vergence power to solve this object distance problem.
We know that any object in front of a lens will produce an image. If object is positioned further than optical infinity (20 feet or 6.0m, or more) this incoming object ray is deemed to be parallel to axis before it strikes the lens, and has a corresponding vergence power of zero (0)D.. ie. an object 2000m from a lens has an incoming vergence in Diopers
L = 1/-2000m= -0.0005D, or so small or negligible that we deem it to have zero (0) vergence power, and thus say that it runs parallel to axis.

An incoming ray object has a vergence power in Diopers that is inversely proportional to its object distance i.e the smaller the object distance l to lens, the larger its incoming minus (-) vergence. This incoming vergence Dioptric power is calculated as
L = n/-l , or L = 1/-l when ray is traveling in air.

Note that this incoming ray vergence power in Diopters of vergence power is written as a minus vergence L = 1/-l(m) because any object distance in front of lens is minus (-) by convention. Although l is the object distance or an actual length, and no length can be a minus length, this close object distance on optical axis produces a minus vergence power that strikes the lens F. A diverging ray is associated with a minus lens power, and a converging ray is associated with a plus vergence power.
So inside 20 feet, an object on axis is always producing a diverging (minus) ray before it strikes the lens, so it produces this minus vergence power L in Diopters.

By conversion, 1m = 100cm = 1000mm.

In this problem object distance l is given as 60cm, or 600mm or 0.6meters, and our incoming vergence power
L = 1/-l from object and can be written as

L = 1(1)/-l(m) = 1/-0.6(m) = -1.67D
L = 1 (100)/- l(cm) = 100/-60(cm) = -1.67D or
L = 1(1000)/-1mm = 1000/-600mm = -1.67D

1/l(m) + 1/f(m) = 1/l’(m) uses its unit of measurement in meters. That is why we consider it important to include units of measurements in any optical formula. Incoming vergence power can be written as L =1/-l(m) = 100/-l(cm) = 1000/-l(mm)

Our F = +4.00D lens has a focal length in meters f(m) = n/F = 1/+4.00 = +0.25meters and in centimeters, this +4.00 lens has the same focal length f(cm) = 100/+4.00 = +25cm or +0.25m. Converting Lens power F = 1/f(m) = 100/f(cm) = 1000/f(mm) and minimize the possibility of decimal error in conversion from cm or mm, to meters.

In this Georgian College video, the instructor substituted for +4.00F = 1/f(m) the number 1/25cm rather than 100/25cm. But units of measurements in this Lensmakers Equation is by convention to be meters (m).

L = 1/-l is the incoming vergence of the object, and 1/-0.06 is not the same as 1/-60.

All of our knowledge in optics is based on vergenge theory and it is fundamental to our knowledge as opticians.

But I too don’t want to argue anymore,either.

Let’s just say that we agree, to disagree. All schools are considered to be autonomous to a certain degree.
Let Georgian College teach opticianry in their own way, and we can leave BC College of Optics to teach opticianry in our own proprietary way.

[MakeOptics]

I will try one more time.
Perhaps I should not have said we “need” to use vergence, but rather that you “can” use vergence power to solve this object distance problem.

That statement makes it an opinion in which case I agree we should use diopter as the unit of measure in the formula, not a ratio as is which is the only description for her method. It's a preference but I agree I prefer diopters and using either meters or mm is my preference but I would use the same formulas you outlined with 1000/l, etc. for the base formula to ensure easy conversion to diopters downstream, if it becomes necessary. I would also assume based on the audience sticking to convention would be a better method to teaching the concept to eliminate confusion. I wouldn't call it wrong necessarily but unconventional, however given the statement we "need" to use vergence if specified in the question or I guess should be specified int he question then it is the wrong way to teach it.