# Uncertainty in the measurement of the focal length in this experiment

• songoku
In summary: Delta f_u=\frac{\partial f}{\partial u} \Delta u$$Similarly, the change in ##f## if you move in the ##v##-direction is$$\Delta f_v=\frac{\partial f}{\partial v} \Delta v$$The square of the total change in ##f## if you move in both directions is$$\Delta f^2=\frac{\partial f}{\partial u}^2 \Delta u^2+\frac{\partial f}{\partial v}^2 \Delta v^2$$If you move in the ##u##-direction by the same songoku Homework Statement The equation connecting object distance u, image distance v and focal length f for a lens is 1/u +1/v = 1/f . A student measures values of u and v, with their associated uncertainties. These are u = 50 mm ± 3 mm and v = 200 mm ± 5 mm. He calculates the value of f as 40 mm. What is the uncertainty in this value? A ±2.1 mm B ±3.4 mm C ±4.5 mm D ±6.8 mm Relevant Equations Percentage and absolute uncertainty ##\frac{1}{f}=\frac{1}{u}+\frac{1}{v}## ##f = \frac{u.v}{u+v}## Let: u + v = w → w = 250 mm ± 8 mm Percentage uncertainy of v = (3/50) . 100% = 6% Percentage uncertainty of u = (5/200) . 100%=2.5% Percentage uncertainty of w = (8/250) . 100% = 3.2% Percentage uncertainty of f = 6 % + 2.5 % + 3.2 % = 11.7 % Absolute uncertainty of f = 11.7 % x 40 = 4.68 mm But the answer should be A Where is my mistake? Thanks songoku said: Homework Statement:: The equation connecting object distance u, image distance v and focal length f for a lens is 1/u +1/v = 1/f . A student measures values of u and v, with their associated uncertainties. These are u = 50 mm ± 3 mm and v = 200 mm ± 5 mm. He calculates the value of f as 40 mm. What is the uncertainty in this value? A ±2.1 mm B ±3.4 mm C ±4.5 mm D ±6.8 mm Relevant Equations:: Percentage and absolute uncertainty Where is my mistake? You are double counting some of the uncertainty. E.g. when u is at the top end of its range, you are adding to f both in regards to its effect on the numerator and its effect on the denominator, but these two effects act oppositely on f. Instead, calculate f from the two highest values of u and v, then do it again for the two lowest values. songoku haruspex said: You are double counting some of the uncertainty. E.g. when u is at the top end of its range, you are adding to f both in regards to its effect on the numerator and its effect on the denominator, but these two effects act oppositely on f. Instead, calculate f from the two highest values of u and v, then do it again for the two lowest values. I think I understand you explanation. Thank you very much haruspex songoku said: I think I understand you explanation. Thank you very much haruspex I should clarify that, in general, for a complicated function of several variables, the safe technique would be to calculate the result for every combination of extreme values the variables (in this case, max u with min v and vice versa), and further to consider whether there may be local extrema of f, where a parameter is somewhere within its range. songoku ## df = \frac {\partial f} {\partial u}~ du + \frac {\partial f} {\partial v}~ dv ## and using root-sum-square error propagation, ## df^2 = ( \frac {\partial f} {\partial u}~ du)^2 + (\frac {\partial f} {\partial v}~ dv)^2 ## I got df = 1.93mm. songoku haruspex said: (in this case, max u with min v and vice versa) Shouldn't we use max u with max v and min u and min v in this case? and further to consider whether there may be local extrema of f, where a parameter is somewhere within its range. I am not sure about this one. We should use differentiation to find local extrema of f? rude man said: ## df = \frac {\partial f} {\partial u}~ du + \frac {\partial f} {\partial v}~ dv ## and using root-sum-square error propagation, ## df^2 = ( \frac {\partial f} {\partial u}~ du)^2 + (\frac {\partial f} {\partial v}~ dv)^2 ## I got df = 1.93mm. I am really sorry I do not understand this method Thanks songoku said: Shouldn't we use max u with max v and min u and min v in this case? Ahem... yes. songoku said: I am really sorry I do not understand this method It is a standard statistical method. It is based on the idea that you would be unlucky to hit the worst combination of u and v errors. It effectively supposes the errors in u and v are normally, independently and identically distributed around 0. Whether that is valid depends on how u and v are measured. E,g. if measured to the nearest mm with a ruler, it might be more accurate to assume a uniform distribution. songoku songoku said: Shouldn't we use max u with max v and min u and min v in this case?I am not sure about this one. We should use differentiation to find local extrema of f?I am really sorry I do not understand this method Thanks You will probbaly soon learn that that is the standard method of error propagation. Particularly with numerous terms in a complex function, doing worst-case analysis is impractical and gives a very unlikely result. And the means of measurement makes no difference. songoku rude man said: You will probbaly soon learn that that is the standard method of error propagation. Particularly with numerous terms in a complex function, doing worst-case analysis is impractical and gives a very unlikely result. And the means of measurement makes no difference. That is all true, at least for scientific work. But I find that it seems some question setters expect the worst case method to be used and others the statistical method, and in most cases the it is not even explained that another approach exists. So it is hard to know, without further info, what the student is expected to do. Another caveat is that engineers tend to use the worst case method. If two components are being made independently to given tolerances then there must be no risk that complying components still do not work together. songoku You get answer (A) by finding the minimum and maximum values that ##f## can have:$$\frac{1}{f_{min}}=\frac{1}{u-\Delta u}+\frac{1}{v-\Delta v}\frac{1}{f_{max}}=\frac{1}{u+\Delta u}+\frac{1}{v+\Delta v}$$Then take the difference and divide by two to put the measured value in the middle between the two extreme values. This method is a "close enough" estimate in most cases of the statistically more correct method suggested by @rude man in post #5 which I will attempt to explain. You have a function of two parameters ##f(u,v)##. You can imagine the measured value ##f(u_0,v_0)## as a point in parameter space, ##\{u_0,v_0\}##, with Cartesian axes ##u## and ##v##. Now if you move only in the ##u##-direction by amount ##\Delta u##, the value of ##f## at that point will be$$f(u+\Delta u,v)=f(u_0,v_0)+\left .\frac{\partial f}{\partial u}\right |_{u=u_0} \Delta u$$which means that the change in ##f## if you move in the ##u##-direction is$$\Delta f_u=\frac{\partial f}{\partial u} \Delta u.$$Likewise if you move in the ##v##-direction only, you have a change$$\Delta f_v=\frac{\partial f}{\partial v} \Delta v.$$These two changes define a ##\Delta f_u\times \Delta f_v## rectangle of uncertainty in the ##u\text{-}v## parameter space within which we expect the value of ##f## to be. A measure of this rectangle is a measure of the uncertainty in ##f##. We use as measure the diagonal of the rectangle (just like the diagonal of a television set is a measure of its size) and write$$\Delta f=\sqrt{(\Delta f_u)^2+(\Delta f_v)^2}=\sqrt{\left( \frac{\partial f}{\partial u} \Delta u\right)^2 +\left( \frac{\partial f}{\partial v} \Delta v\right)^2 }$$which is what @rude man has (except for the radical) in post #5. Once you understand how it works with two parameters, it is easy to see how it can be expanded to ##N## parameters in ##N##-parameter space with an ##N##-dimensional box of uncertainty. Taking derivatives is much more convenient than using the min-max method especially when you have non-linear functions such as sinusoidals, exponentials and inverse trig functions all jumbled together. Last edited: songoku and hutchphd kuruman said: You get answer (A) by finding the minimum and maximum values that ##f## can have:$$\frac{1}{f_{min}}=\frac{1}{u-\Delta u}+\frac{1}{v-\Delta v}\frac{1}{f_{max}}=\frac{1}{u+\Delta u}+\frac{1}{v+\Delta v}$$Then take the difference and divide by two to put the measured value in the middle between the two extreme values. This method is a "close enough" estimate in most cases of the statistically more correct method suggested by @rude man in post #5 which I will attempt to explain. You have a function of two parameters ##f(u,v)##. You can imagine the measured value ##f(u_0,v_0)## as a point in parameter space, ##\{u_0,v_0\}##, with Cartesian axes ##u## and ##v##. Now if you move only in the ##u##-direction by amount ##\Delta u##, the value of ##f## at that point will be$$f(u+\Delta u,v)=f(u_0,v_0)+\left .\frac{\partial f}{\partial u}\right |_{u=u_0} \Delta u$$which means that the change in ##f## if you move in the ##u##-direction is$$\Delta f_u=\frac{\partial f}{\partial u} \Delta u.$$Likewise if you move in the ##v##-direction only, you have a change$$\Delta f_v=\frac{\partial f}{\partial v} \Delta v.$$These two changes define a ##\Delta f_u\times \Delta f_v## rectangle of uncertainty in the ##u\text{-}v## parameter space within which we expect the value of ##f## to be. A measure of this rectangle is a measure of the uncertainty in ##f##. We use as measure the diagonal of the rectangle (just like the diagonal of a television set is a measure of its size) and write$$\Delta f=\sqrt{(\Delta f_u)^2+(\Delta f_v)^2}=\sqrt{\left( \frac{\partial f}{\partial u} \Delta u\right)^2 +\left( \frac{\partial f}{\partial v} \Delta v\right)^2 }which is what @rude man has (except for the radical) in post #5.

Once you understand how it works with two parameters, it is easy to see how it can be expanded to ##N## parameters in ##N##-parameter space with an ##N##-dimensional box of uncertainty. Taking derivatives is much more convenient than using the min-max method especially when you have non-linear functions such as sinusoidals, exponentials and inverse trig functions all jumbled together.
Yes.

As an engineer I can firmly state that the statistical method is the only one I ever used.

The approach is to use that method, then determine if the product meets specs. Occasionally it may be necessary to make final adjustments, but that is seldom encountered. It is very much more cost-effective to go statistical and check for ultimate compliance than it is to use the awkward and usualy extremely punishing worst-case method unless only 1 or 2 terms are involved.

songoku
rude man said:
Yes.

As an engineer I can firmly state that the statistical method is the only one I ever used.

The approach is to use that method, then determine if the product meets specs. Occasionally it may be necessary to make final adjustments, but that is seldom encountered. It is very much more cost-effective to go statistical and check for ultimate compliance than it is to use the awkward and usualy extremely punishing worst-case method unless only 1 or 2 terms are involved.
I'm with you. However, if one is a teacher of students who have never seen partials or even calculus, one waves one's hands and uses a method that is easy to explain in terms of hi-lo discrepancy. It may not be appropriate for complicated equations, but then again students who have not seen calculus are not likely to encounter many of these.

songoku
kuruman said:
I'm with you. However, if one is a teacher of students who have never seen partials or even calculus, one waves one's hands and uses a method that is easy to explain in terms of hi-lo discrepancy. It may not be appropriate for complicated equations, but then again students who have not seen calculus are not likely to encounter many of these.
Yes, but you usually don't need calculus to do root-sum-squares analysis. Just a few rules applying to sums & differences, percentages of products and quotients, and "n times powers of n", to name the ones that should be memorized.

Of course, as you say, when measurements like sines and cosines are involved you have to take derivatives. Burt then ...

songoku
I get the same result (19.3 mm) using the method explained. Thank you very much for all the help and explanation haruspex, rude man, kuruman

## 1. What is uncertainty in the measurement of focal length?

Uncertainty in the measurement of focal length refers to the range of values within which the true value of the focal length may lie. It is a measure of the precision and accuracy of the measurement.

## 2. How is uncertainty calculated in the measurement of focal length?

Uncertainty in the measurement of focal length is typically calculated using statistical methods, such as standard deviation or confidence intervals. It takes into account factors such as instrument precision, human error, and variability in the measurement process.

## 3. What factors can contribute to uncertainty in the measurement of focal length?

There are several factors that can contribute to uncertainty in the measurement of focal length, including the precision of the measuring instrument, the skill and technique of the person taking the measurement, and any external factors that may affect the measurement, such as environmental conditions.

## 4. How can uncertainty in the measurement of focal length be reduced?

Uncertainty in the measurement of focal length can be reduced by using more precise measuring instruments, improving measurement techniques, and taking multiple measurements and calculating an average. It is also important to control for any external factors that may affect the measurement.

## 5. Why is it important to consider uncertainty in the measurement of focal length?

Considering uncertainty in the measurement of focal length is important because it provides a measure of the reliability and accuracy of the measurement. It also allows for a better understanding of the limitations of the measurement and helps to determine the level of confidence in the results.

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