Can viewing distance be used to measure eye strain?

[Miopen]

I am doing independent research to compute eye strain and I hit a problem. Hope someone can help.

This is the situation.

Work done to pull a spring is ½kx².

The human lens is like a spring. Linear, and obeys Hooke’s law.

Work done to stretch the lens is also ½ kx² (k=spring constant of lens, x = lens stretch)

Since stretching of the lens causes eye strain, I created an equation:

Eye strain = work done = ½ k x²

So far so good. What follow is tricky.

The minute stretching of the lens correlates to change in viewing distance in real life but in reverse order, i.e. lens stretch increases, viewing distance decreases.

Instead of measuring lens stretch to compute eye strain, which is almost impossible, I want to measure viewing distance to compute eye strain.

If the change in viewing distance in real life is (d)

Eye strain = work done = ½ k (1/d²)

Question 1: Can I legally inverse (d) to reflect viewing distance is the inverse of lens stretch (x)? (Looking at the equation, it looks correct. Eye strain is the inverse of distance focus.)

Question 2: The SI unit for work done is N/m. This is not the case after I inverse (d). It becomes N/m³ . How do I resolve the SI units or do I need to?

Thanks for helping.

 

[jbriggs444]

Question 1: Can I legally inverse (d) to reflect viewing distance is the inverse of lens stretch (x)? (Looking at the equation, it looks correct. Eye strain is the inverse of distance focus.)

At what distance does a relaxed lens focus? You may have neglected a term in your equation.
Is the change in focal length inversely proportional to the deflection of the lens? You assume so, but is it true?
If so, what is the constant of proportionality and what units does it have?

Question 2: The SI unit for work done is N/m. This is not the case after I inverse (d). It becomes N/m³ . How do I resolve the SI units or do I need to?

The SI unit for work done is N-m. Newtons multiplied by meters. In addition, the constant of proportionality for focal length in inverse proportion to deflection will have units on it.

 

[Dale]

Question 1: Can I legally inverse (d) to reflect viewing distance is the inverse of lens stretch (x)? (Looking at the equation, it looks correct. Eye strain is the inverse of distance focus.)

What you need to do is formally write the expression relating x and d. Then solve that expression for x. Once you have done that then you can legally substitute that expression into the expression for work.

As you have written it the expression is $x=1/d$ which is clearly wrong because it is dimensionally inconsistent (as you observed).

 

[DCadorin]

It looks like that more relevant is the distance from the lens to the image than the one of the object to the lens for your calculations.

Extracted from wikipedia that might help:
For a thin lens in air, the focal length is the distance from the center of the lens to the principal foci (or focal points) of the lens. For a converging lens (for example a convex lens), the focal length is positive, and is the distance at which a beam of collimated light will be focused to a single spot. For a diverging lens (for example a concave lens), the focal length is negative, and is the distance to the point from which a collimated beam appears to be diverging after passing through the lens.

When a lens is used to form an image of some object, the distance from the object to the lens u, the distance from the lens to the image v, and the focal length f are related by

1/f=1/u+1/v

https://wikimedia.org/api/rest_v1/media/math/render/svg/3667a4f96bd5863e1a007aa89dcc0efc1abed2c7
The focal length of a thin lens can be easily measured by using it to form an image of a distant light source on a screen. The lens is moved until a sharp image is formed on the screen. In this case 1/u is negligible, and the focal length is then given by

f almost equal to v

https://wikimedia.org/api/rest_v1/media/math/render/svg/e944e4ed0cd41dcd8e97ca65a147b3bf0fadaf8e

 

[jbriggs444]

It looks like that more relevant is the distance from the lens to the image than the one of the object to the lens for your calculations.

The distance from the lens to the image (hopefully in focus on the retina) is a constant. The muscles responsible for maintaining focus do not do so by moving the lens toward or away from the retina. They do so by deforming the lens, altering its focal length.

 

[jim mcnamara]

Here is a clear discussion of accommodation - the eye changing focus: https://en.wikipedia.org/wiki/Accommodation_(eye)
@jbriggs444 explanation is clear but pictures always help

 

[Miopen]

I have to read up on optics to understand your reply. It turns out that my question is not be related to optics.
I will provide more information so that you understand how I come to the question.

Elasticity is a physical property of material. The lens in our eye is elastic, and I only need to deal with the elastic range of the lens.
During accommodation, the lens is stretched as seen in the picture below from Wiki link (Jim McNamara.).

The force to stretch the lens by a minute distance x is given by this formula:

F = k x

Where F is the force provided by ciliary muscle, and k is the lens’ spring constant (SI: N/m)

We can see that the lens at the bottom picture becomes thicker when the viewing distance decreases. If the viewing distance is D, then x is inversely proportional to D

x = 1/D

<---------D------>

The work done to stretch the lens is given by this formula (because the lens behaves like a spring):

work done = 1/2 k x²

Work done is the same as energy used. When we use energy to lift a heavy object, we feel strain in our muscles. Same goes for the ciliary muscle when it stretches the lens on accommodation; and we get eye strain. So,

work done = energy used = eye strain = 1/2 k x² ------------ (a)

I do not know how to change the equation when I extrapolate the stretching of lens x to the decrease in viewing distance D?

I want to stress that I do not have to compute the actual work done on the lens, which is measured in dyne/mm. I only want a COMPARATIVE figure of eye strain on a scale in meters.

I can think of three possibilities:

Eye strain = 1/2 k (1/D²) --------- (b) Nm = N/m³

Eye strain = 1/(1/2 k D²) --------- (c) Nm = 1/(Nm)

1/Eye strain = 1/(1/2 k D²) --------- (d) 1/(Nm) = 1/(Nm)

Q1. If I am not concern with actual measurement, do I need to worry about balancing SI on both side of the equation?

Q2. Which is the correct equation to use?

Thanks for helping.

 

[jbriggs444]

Work done is the same as energy used

Not when using human muscles, it isn't. If you've ever held up a gallon jug at arm's length, you will quickly notice the discrepancy between work done (zero) and energy required (decidedly non-zero).

 

[Dale]

If the viewing distance is D, then x is inversely proportional to D

x = 1/D

<---------D------>

What do you think of this equation? What are the units of x and D. Is this equation dimensionally consistent?

 

[Miopen]

What do you think of this equation? What are the units of x and D. Is this equation dimensionally consistent?

I am not good in physics and math but I will try.

x is in mm, D is in meter. Depending on age, x can be stretched to around 1 mm, and D, the near point or amplitude of accommodation 10-15 cm.
When x is zero (lens relaxed), D is infinity (or 6 m is as good as infinity when rays entering the eyes are parallel).

Dimensionally, m <> 1/m. I am not sure how to resolve this. Please explain.

 

[Dale]

Dimensionally, m <> 1/m. I am not sure how to resolve this. Please explain.

That is a big clue that the equation is wrong. There must be something else, such as a constant of proportionality. E.g. Maybe the correct equation is $x=k/D$ where k is some factor in units of m^2.

Edit: k in this equation would not be the same as k in Hooke's law, I probably should have used a different letter, maybe K.

 

[Miopen]

Now I get it.

Let P be the constant of proportionality, then
x = P/D

Assuming x increases by 1mm while D decreases by 6m

0.001 = P / 6

P = 0.001 * 6 = 0.006 (m²)

From equation (a), work done = ½ k x² , substituting x for P/D

work done = ½ k (P/D)² - dimensionally Nm = Nm^-1 (m⁴ / m²) = Nm

The constant of proportionality needs verification. For the equation, I think I get it right this time.

Thank you very much.

 

[Miopen]

Not when using human muscles, it isn't. If you've ever held up a gallon jug at arm's length, you will quickly notice the discrepancy between work done (zero) and energy required (decidedly non-zero).

You are right. A classical question “why does holding something up cost energy while no work is being done?”

If I insist on measuring the amount of energy used, not by work done on the object, not by thermodynamics, with addition of time factor, can this be done?

I will change the question into linear form involving a spring:

A force is used to extend a spring by 10 cm. The spring constant is 260 N/m

Work done to pull the spring:

Work = ½ k x^2 = ½ * 260 * (0.1)^2 = 1.3 Nm

My hand supply the energy to pull the spring. 1.3 Nm of energy is used.

If I hold on to the spring and keep it stretched at 10 cm for 10 minutes, How do I calculate the total energy used? Can I multiply time into energy used?

Total energy used = 1.3 * (10 x 60 sec) = 780 Nms (Joule s)

Is this legitimate?

 

[jbriggs444]

If I insist on measuring the amount of energy used, not by work done on the object, not by thermodynamics, with addition of time factor, can this be done?

No, there is no way to deduce the answer from the basic principles of physics. It is a medical question and requires knowledge of the mechanics of the ciliary muscles.