Transposition problem

[sevalav]

Can someone help me?

This prescription we have: +8,00 / +3,00 x 90
Transposition this prescription in toric form (with spherical curve +12,00) is :
+12,00 .
- 1,00 x 90 / - 4,00 x 180 (thin lens)

But if we must produced this lens with 10mm thickness (t), with n=1,5, BVP in this situation is
+12,04 x 90 / +90,04 x 180 (or +9.04 / +3.00 x 90 - sfero-cyl. form. )
We must now reduce high power of front curve, if we wont right BVP. Below is the method and results (in ax 90°):
L1.................. ....... - 11,00
F1........................... - 1,00
n............................ 1,50
t............................. 0,01m
V= L1 + F1 .............. -12,00
L2= V/(1-t/n * V)... -11,11
F2 = - L2 ............... +11,11

My question, for start:
how they get L1 (-11,00)?

[Happylady]

???

[sevalav]

This case is from my school. But I do not understood this case in total. Please, if someone can describe this case, please describe it for me.
Im sorry for my bad english.

[HarryChiling]

Can someone help me?

This prescription we have: +8,00 / +3,00 x 90
Transposition this prescription in toric form (with spherical curve +12,00) is :
+12,00 .
- 1,00 x 90 / - 4,00 x 180 (thin lens)

But if we must produced this lens with 10mm thickness (t), with n=1,5, BVP in this situation is
+12,04 x 90 / +90,04 x 180 (or +9.04 / +3.00 x 90 - sfero-cyl. form. )
We must now reduce high power of front curve, if we wont right BVP. Below is the method and results (in ax 90°):
L1.................. ....... - 11,00
F1........................... - 1,00
n............................ 1,50
t............................. 0,01m
V= L1 + F1 .............. -12,00
L2= V/(1-t/n * V)... -11,11
F2 = - L2 ............... +11,11

My question, for start:

how they get L1 (-11,00)?

I am assuming that your variable is going to be the front curve and you are looking for a way to work the step along in reverse. If that's the case then:

In the step along

L1 = 0.00
F1 = ?.??
n = 1.50
t = 0.01
L2 = (L1 + F1)/(1-t/n * (L1 + F1))
F2 = -1.00
BVP = L2 + F2 = +11.00

That's the step along equation, you were missing the component of the incident power on the back surface.

To rework it

BVP = L2 + (-1.00)
+11.00 = L2 + (-1.00)
L2 = +12.00

So now we have our incident rays on the back surface, Lets work our way up one more step

L2 = (L1 + F1)/(1-t/n * (L1 + F1))
+12.00 = (F1) / (1-0.01/1.50 * F1)
F1 = +12.00 (1-0.01/1.50 * F1)
F1 = +12.00 - 0.08 * F1
+12.00 = F1 + 0.08 * F1
+12.00 = 1.08 * F1
F1 = +11.11D

Hope that helps, I make a lot of little math mistakes but the theory is solid, just double check it to make sure.

[sevalav]

Lets start again.

This case was represent to me in this way:

A) The lens (+8,00 DS / +3,00 DC x90) in spherical-cyl. form convert to toroidal form as a thin lens, with base curve +12,00 DS (front surface).

B) Calculate BVP (on ax 90) for previously lens (+8,00 DS / +3,00 DC x90), if lens have center thickness t = 0,01m and n = 1,5. Back surface is toroidal.

Answer
A): +8,00 DS / +3,00 DC x90 (front) base curve: +12,00 DS

1.+8,00 DC x 180 / +8,00 DC x 90
......................... +3,00 DC x90
---------------------------------------
+8,00DC x 180 / +11,00 DC x90

2.+8,00 - (+12,00) = -4,00 DC x 180
3.+11,00 - (+12,00) = -1,00 x 90

4.as a result, we have lens in toroidal form:
................+12,00 DS
--------------------------------------
-1,00 DC x90 / -4,00 DC x180

For me, there is no problem to understood case under A). I wrote all steps because the B) need same data from A).

B) In book, case under B) start with this data:
t = 0,01m
n = 1,50
F1 = -1,00 D
L1 = -11,00 D
F2=?

Also, book mentioned that the lens is now reversed (look picture) and in this case we must do that because, if we wont computing BVP, we need far distance object position. This explanation are not enough for me.

1.In this point, can you explain why we must reserve lens in this case?

2. why is used 11,00 D for L1 ? Because this power is total lens power on 90° (look cross cyl form in A) 1.)?

3.Is the reason for minus (-) before 11,00 because L1 (ex L'2) is now on left side from front surface (before back surface)?

[sevalav]

Picture

[HarryChiling]

Lets start again.

This case was represent to me in this way:

A) The lens (+8,00 DS / +3,00 DC x90) in spherical-cyl. form convert to toroidal form as a thin lens, with base curve +12,00 DS (front surface).

B) Calculate BVP (on ax 90) for previously lens (+8,00 DS / +3,00 DC x90), if lens have center thickness t = 0,01m and n = 1,5. Back surface is toroidal.

Answer
A): +8,00 DS / +3,00 DC x90 (front) base curve: +12,00 DS

1.+8,00 DC x 180 / +8,00 DC x 90
......................... +3,00 DC x90
---------------------------------------
+8,00DC x 180 / +11,00 DC x90

2.+8,00 - (+12,00) = -4,00 DC x 180
3.+11,00 - (+12,00) = -1,00 x 90

4.as a result, we have lens in toroidal form:
................+12,00 DS
--------------------------------------
-1,00 DC x90 / -4,00 DC x180

For me, there is no problem to understood case under A). I wrote all steps because the B) need same data from A).

B) In book, case under B) start with this data:
t = 0,01m
n = 1,50
F1 = -1,00 D
L1 = -11,00 D
F2=?

Also, book mentioned that the lens is now reversed (look picture) and in this case we must do that because, if we wont computing BVP, we need far distance object position. This explanation are not enough for me.

1.In this point, can you explain why we must reserve lens in this case?

2. why is used 11,00 D for L1 ? Because this power is total lens power on 90° (look cross cyl form in A) 1.)?

3.Is the reason for minus (-) before 11,00 because L1 (ex L'2) is now on left side from front surface (before back surface)?

I don't really undrstand what you are saying, and I am afraid that my serbian is worst than your english so it may be a communication issue here, but I would be more than happy to help. Can you scan the book or type it exactly as it appears in your text, this may help eliminate any confusion.

[lensgrinder]

Lets start again.

This case was represent to me in this way:

A) The lens (+8,00 DS / +3,00 DC x90) in spherical-cyl. form convert to toroidal form as a thin lens, with base curve +12,00 DS (front surface).

B) Calculate BVP (on ax 90) for previously lens (+8,00 DS / +3,00 DC x90), if lens have center thickness t = 0,01m and n = 1,5. Back surface is toroidal.

Answer
A): +8,00 DS / +3,00 DC x90 (front) base curve: +12,00 DS

1.+8,00 DC x 180 / +8,00 DC x 90
......................... +3,00 DC x90
---------------------------------------
+8,00DC x 180 / +11,00 DC x90

2.+8,00 - (+12,00) = -4,00 DC x 180
3.+11,00 - (+12,00) = -1,00 x 90

4.as a result, we have lens in toroidal form:
................+12,00 DS
--------------------------------------
-1,00 DC x90 / -4,00 DC x180

For me, there is no problem to understood case under A). I wrote all steps because the B) need same data from A).

B) In book, case under B) start with this data:
t = 0,01m
n = 1,50
F1 = -1,00 D
L1 = -11,00 D
F2=?

Also, book mentioned that the lens is now reversed (look picture) and in this case we must do that because, if we wont computing BVP, we need far distance object position. This explanation are not enough for me.

1.In this point, can you explain why we must reserve lens in this case?

2. why is used 11,00 D for L1 ? Because this power is total lens power on 90° (look cross cyl form in A) 1.)?

3.Is the reason for minus (-) before 11,00 because L1 (ex L'2) is now on left side from front surface (before back surface)?

OK, I am going to make a few assumptions here. The vergenge of light entering the lens is -11.00 D (L1) and based on the image you provided we want the the light to exit with 0 vergence (L'2)


You reverse the lens to show what happens to a lens when we look through the other surface.

L1 is the vergence of light and is negative because it is diverging from its source. We generally will set L1 to 0 because we are dealing with parallel light incident on the surface.

[HarryChiling]

OK, I am going to make a few assumptions here. The vergenge of light entering the lens is -11.00 D (L1) and based on the image you provided we want the the light to exit with 0 vergence (L'2)


You reverse the lens to show what happens to a lens when we look through the other surface.

L1 is the vergence of light and is negative because it is diverging from its source. We generally will set L1 to 0 because we are dealing with parallel light incident on the surface.

Awesome post, The following line threw me:

We must now reduce high power of front curve, if we wont right BVP. Below is the method and results (in ax 90°):

I didn't even realize he had posted an image of the problem.

[lensgrinder]

Thanks Harry. I had to read it more than once too. The negative vergence threw me. Normally we deal with 0 vergence.

[sevalav]

Thanks a lot! Does someone know some place on net where I can find the math problem like this, for exercise?

[JRS]

You might try some of these Sevalav. Unfortunately, I could only find english language versions.

http://www.opticampus.com/

http://www.opticalformulastutorial.com/

http://opticalinstructor.homestead.com/

[sevalav]

Thanks JRS

[Darryl Meister]

I would say that your book is over-complicating the matter. Since we already know the front curve (12.00 D), and we are considering the power of the lens for an infinite object (that is, the back vertex power), the vergence equation and L₁ (equal to 0 in this case) variable are unnecessary.

Ultimately, they are trying to demonstrate the difference in power between an infinitely "thin" lens and a "real" lens of finite thickness.

For a "thin" lens, the back curve can be determined by simply subtracting the front curve F₁ from the desired back vertex power. The back vertex power F_V of a "thin" lens is given by:



So the back curve F₂ is given by:



For a thick lens, however, you must consider the effects of center thickness, which serves to increase the effective power produced by the front surface. In this case, the back vertex power F_V of a "thick" lens is given by:



So the back curve F₂ is given by:



These formulas can be used for each of the principal meridians of the lens in order to determine the two back curves (that is, +8.00 D and +11.00 D in your example). Alternatively, if the lens is in minus-cylinder form, you can determine the back curve of the meridian with the most plus power and then just add the cylinder power to it to arrive at the second curve.

[Hermi Hidalgo]

You might try some of these Sevalav. Unfortunately, I could only find english language versions.

http://www.opticampus.com/

http://www.opticalformulastutorial.com/

http://opticalinstructor.homestead.com/

J.R.

I am new to this forum and I found these websites to be very helpful.

Thanks very much.

Hermi