Thick edge negative meniscus exam calculation

[marcuslucas]

Hi everyone,
I'm new to the thread but have found it to be an invaluable learning resource thus far. I was hoping to get some insight into
an exam question example.
A -4.00D lens is made in glass of refractive index 1.53 with a +5.00D base curve and edged to a 46x40 oval shape. The thin-edge substance of the lens is 3.0mm. what is the thick-edge substance?

What I've done so far is to calculate the the front radius as 106mm and the back surface radius as 58.88m, from front surface +5.00D and back surface -9.00D . Then using the exact sag formula I have calculated four sag values; two in the horizontal meridian a diameter of 23mm and two in the vertical meridian of 20mm
s1 = 2.52mm , s2= 4.67mm ,s3 = 1.09mm , s4 = 3.50mm. This is the part i'm having trouble with I don't know how to proceed from here any help or suggestions would be greatly appreciated.

Thanks in advance.

[tmorse]

Now use the Edge/Centre thickness formula...
Tc-Tp = s1-s2
where Tc= Centre thickness (mm)
Tp= peripheral (edge) thickness (mm)
s1= sag of front surface (mm)
s2= sag of back surfdace (mm)

I suggest you purchase Ellen Stoner's Math Formula Tutorial textbook.

[MakeOptics]

Hi everyone,
I'm new to the thread but have found it to be an invaluable learning resource thus far. I was hoping to get some insight into
an exam question example.
A -4.00D lens is made in glass of refractive index 1.53 with a +5.00D base curve and edged to a 46x40 oval shape. The thin-edge substance of the lens is 3.0mm. what is the thick-edge substance?

What I've done so far is to calculate the the front radius as 106mm and the back surface radius as 58.88m, from front surface +5.00D and back surface -9.00D . Then using the exact sag formula I have calculated four sag values; two in the horizontal meridian a diameter of 23mm and two in the vertical meridian of 20mm
s1 = 2.52mm , s2= 4.67mm ,s3 = 1.09mm , s4 = 3.50mm. This is the part i'm having trouble with I don't know how to proceed from here any help or suggestions would be greatly appreciated.

Thanks in advance.

So you have figured out the curves using the fomula:

Power = front curve + back curve

-4.00 = +5.00 + (-9.00)

Then you find the radius of both curves:

Front = (.53)/5 = 106mm
Back = (.53)/9 = 58.8mm

Now the shape is an oval with a diameter of 46x40, so sag the front and back of both meridians:

sag(front)@46mm = 106 - sqrt(106^2 - 23^2)
= 106 - sqrt(11236 - 529)
= 106 - sqrt(10707)
= 106 - 103.5 = 2.5mm
sag(front)@40mm = 106 - sqrt(11236 - 400)
= 106 - sqrt(10836)
= 106 - 104.1 = 1.9mm
sag(back)@46mm = 58.8 - sqrt(58.8^2 - 23^2)
= 58.8 - sqrt(3457.44 - 529)
= 58.8 - sqrt(2928.44)
= 58.8 - 54.1 = 4.7mm
sag(back)@40mm = 58.8 - sqrt(3457.44 - 400)
= 58.8 - sqrt(3057.44)
= 58.8 - 55.3 = 3.5mm

Now we have to figure out the thicknesses in both meridians:

thick@46mm = 4.7 - 2.5 = 2.2mm
thick@40mm = 3.5 - 1.9 = 1.6mm

So now we know our thin edge is along the 40mm diameter side of the lens and since our give was that the thin edge substance is 3.0mm, then the center thickness is:

center thickness = 3.0 - 1.6 = 1.4mm

Add the center thickness to the thick edge and we have our answer:

thick edge substance = 1.4 + 2.2 = 3.6mm

Take the problem and break it into steps, then follow along. I don't like to label things as s1, s2, s3, s4, etc. it has no meaning that way. If you noticed I labeled using more descriptive variables. It is common in coding to label variables in such a way to make the code more readable I find it also works well in optics when a problem needs to be followed through in multiple steps. Good luck.

[marcuslucas]

Thanks for your time, the step by step calculation was excellent, clear and well presented. Many Thanks.