The formula that we will use here is in  Systems for Ophthalmic Dispensing on page 358-9, and in Optical Formulas Tutorial, pages 142-146.  Start this module by reading.

For lens absorption and transmission we have four basic formulas.  Find out from your instructor if:
1. you are required to memorize them; or
2. you just need to know know to use them; or
3. you can skip using the formulas completely.


FORMULAS:

Surface reflection, for a lens that is in air.	(n - 1) (I) (n + 1)
Transmission for a single layer of absorptive material (plastic, coatings)	(q)(I)
Transmission for repeating layers of absorptive material (glass)	(q)(I)
Successive amounts of transmission:  "ultimate transmission"	(T1)(T2)

Where:
n = index of reflection of the material of the lens material
q = the transmission of the lens material, per unit of thickness
x = the number of units of thickness in the lens
I = intensity of light, decimal representation of percent.
(We start with 100%, represented as I = 1.0)
T1 and T2 are amounts of transmission, represented in decimal form

Let's start with that last one first.

EXAMPLE 1
Suppose that you have a light tint in the plastic lens, say, a 10% rose tint.  That means that 10% of the light is absorbed.  Then 100 - 10 = 90% of the light is transmitted.  So, the transmission value of the lens is the decimal form of 90%, which is 0.90.

Now you add a sunglass clip-on to the glasses.  The clip-on has a transmission value of 15%.  In decimal form, that is a transmission of 0.15.

When the glasses wearer puts the clip-on on the glasses the total transmission is (T1)(T2) = (0.90)(0.15) = 0.135 = 13.5% transmission.

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EXAMPLE 2
Let's take that same pair of glasses with the 10% rose tint.  Put the person in a car with a tinted windshield, say, 15% absorption.

Now T1, the glasses transmission, is still 100 - 10 = 90% = 0.90.  T2 = 100 - 15 = 85% = 0.85.

The total transmission for this person when driving the car is (T1)(T2) = (0.90)(0.85) = 0.765 = 76.5%.  Slightly more than 25% of the ambient light is blocked by the two tints.  If this person is driving at night then the ability to see detail in low-illumination areas, like the side of the road, has been decreased by the combination of glasses tint and windshield tint.

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EXAMPLE 3 
Now let's look back at that lens with the 10% rose tint, and see what it's transmission is when we take surface reflections into account.  If you have not had the surface reflections lesson yet you will need to wait on this example until after you have read it through.

Let's make this lens out of polycarbonate, with an index of refraction of 1.586.  We will start with 100% of the ambient light.  If it helps, think of this as 100 rays of light.

First, we have some light reflecting off of the first surface of the lens.  The formula is the first one in the box at the beginning of this lesson:

(n - 1)	(I)
(n + 1)

In this case, since we are starting with 100% of the light, I is 100% = 1.00, and n is 1.586.  The result is that 0.051 = 5.1% of the light is reflected from the front surface.  That is 5.1 of those 100 rays that will not ever enter the lens.  Therefore, 100 - 5.1 = 94.9 rays, or 94.9% = 0.949 of the light will enter the lens.

Now we have the pink tint that is removing 10% of the light.  It's transmission is 0.90.  So, T1 is 0.949 and T2 is 0.9, and (T1)(T2) = (0.949)(0.90) = 0.854 = 85.4% of the light reaches the back surface of the lens.

But 5.1% = 0.051 reflects from each surface.  We had 85.4 of those original rays reach the back;  now another 5.1% will reflect back into the lens.  T1 is now 0.854 and T2 = 0.051, and (T1)(T2) = (0.854)(0.051) = 0.044 = 4.4% more that is not transmitted through the back surface.  85.4 rays reached the back surface, 4.4 of them reflected back inside the lens, so 81.0 rays will get out of the lens through the back surface.

NOW put that person in the car with the tinted windshield driving at night.  Remember, there will be reflections from the car's windshield, too, so that the 85% transmission through the car windshield will also actually be reduced.  T1 is now 0.85 and T2 is 0.81.  The person will receive (0.85)(0.81) = 0.689 or 68.9% or 68.9 rays for every 100 that are available.  That means that now the person is only getting 2/3 of the light available in those low illumination areas by the side of the road . . .

Do you get the suspicion that I do not like to dispense glasses with a light tint to people who will be driving in them?  Do you see one of the reasons why anti-reflective coatings are valuable when driving at night?

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EXAMPLE 4 
Now let's look at a lens with an Rx of -5.00, made from a gray green glass that has a transmission value of 75% for every mm of thickness.  You may have never seen one of these, and you may never see one in a whole career of opticianry, so you may not need to learn to do this example.  You already know a number of reasons why we rarely sell Rx'd glass lenses;  now you will 'see' another one.

We will assume that this lens will have a center thickness of 2.0 mm.  And since the sagittal formula for lens thickness is in another course, I will not make you use it;  I am going to tell you  that the edge thickness for this lens (minimum blank size of 50 mm) will be 5.1 mm.  [If you have taken a course that included the sag formula then I suggest that you verify my edge thickness. . . I used the approximation formula.]

The transmission for the center of this lens is (q)(I), where q (the transmission) is 0.75, x (the thickness) is 2 (since this lens material has a transmission of 75% per mm and the lens is 2 mm thick) and I is the amount of incident light.  Since this is glass, 4.3% reflects from the front surface, so I = 100 - 4.3 = 95.7% of the incident light that enters the lens.  So, (.75)(95.7) = 53.8 or 53.8% of the light reaches the back surface at the center of the lens.  4.3% of this light reflects back inside of the lens, and (0.043)(53.8) = 2.3, so 2.3% of the original light reflects back, and 53.8 - 2.3 = 51.5% of the original light will be transmitted through the lens.

On the edge of the lens, where is is 5.1 mm thick, the transmission is:

Start with 100%
4.3% reflects off the front surface, so 95.7% enters the lens
(0.75)(95.7) = and how do you do this?
If your calculator has sin, cos and tan on it then it has a key that looks like .  That is the amount (x) raised to the power (y).  So you punch in the x, .75, then you punch  and then you punch the y, 5.1 and the = key and your calculator should say 0.23057.... and that is the what 75% raised to the 5.1 power is.  So,
(0.75)(95.7) = (0.23057)(95.7) = 22.1
4.3% of 22.1 reflects back inside the lens, so
(22.1)(0.043) = 0.9, and
22.1 - 0.9 = 21.2% of the original light will exit the lens at the edge.
SO?  The transmission through the center of the lens is 51.5%, or about 1/2 of the light, and the transmission  through the edge of the lens is 21.2%, or less than 1/4 of the light.  SO?  The lens looks a whole lot lighter in the center than at the edges.




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EXAMPLE 5
The Rx is +5.00 -2.00 x 045.  The lens is made from absorptive glass with an absorption rating of 30% per mm. The lens is surfaced with a 1.0 minimum edge thickness at a 55 mm diameter.  What will the lens look like?
 

1. An absorption rating of 30% is a transmission rating of 70%, or 0.70.
2. Thickness:
1. The minimum edge is 2.0 mm.
2. The center will be 4.6 mm.
3. The thick edge (in the meridian where the power is +2.00) will be 2.4 mm.  [Don't ask.]
3. Reflection from the front surface is 4.3%, so 95.7% of the light enters the lens.
4. At the center, where the lens is 4.6 mm thick, the transmission will be
(0.70)(95.7) = (0.1938)(95.6) = 18.5%
4.3% of the 18.5 is 0.8, so 17.7% of the light is transmitted through the center.
5. At the thin edge where the thickness is 1 mm the transmission will be
(0.7)(95.7) = 66.99%
4.3% of 66.99 is 2.9, so 66.99-2.88 = 64.1% of the light is transmitted through the thin edge.
6. At the thick edge where the thickness is 2.4 mm the transmission will be
(0.70)(95.7) = (0.4245)(95.7) = 40.7%
4.3% of 40.7 is 1.7; so 40.7 - 1.7 = 39% of the light is transmitted through the thick edge.
What does this lens look like?  It has a dark center, and the edges at the 40 and 225 directions are moderately light, while the edges at 135 and 315 meridians are in-between.

Photochromatic glass has the same problems, but not to that extreme, because the inner layers in the thick areas do not absorb as much as the outer layers do.  Photochromatic glass has the extra problem, when you are making fused bifocals, in that the segment glass will not be photochromatic.  Since the amount of regular glass behind the segment will be thin, the segment area will be lighter than the rest of the lens.

REFERENCES:
Brooks & Borish, Systems for Ophthalmic Dispensing, 2nd edition Chapter 12 (pages 329 - 364).
Fannin & Grosvenor, Clinical Optics 2nd edition, Butterworth-Heinemann, pages 167 - 195.
Stoner, & Perkins, Optical Formulas Tutorial page 142-149.