# Homework Help: Large Uncertainty Became Unsignificant After Calculations?

1. Jan 15, 2012

### xGary

1. The problem statement, all variables and given/known data

Calculating focal lengths of lenses.

s = 144.7 ± 5.52 mm
s' = 86.0 + 0.71 mm

2. Relevant equations

Focal Length
f = (s)(s') / ( s + s' )

A + B ± root [ (±A)^2 + (±B)^2 ]

Multiplying/Dividing

AB ± root [ (±A/A)^2 + (±B/B)^2) ]

3. The attempt at a solution

I have:

s = 144.7 ± 5.52 mm
s' = 86.0 + 0.71 mm

and am getting:

f = 53.94 ± 0.02412 mm

My question is, does it make sense that the uncertainty that was obviously significant at the start (5.52mm uncertainty in a 144.7mm magnitude) became so miniscule at the end? The way I am calculating this is

1) Calculating numerator and denominator of focal length equation separately
-Top = ( s * s' ) and get uncertainty of 0.039mm using the uncertainty multiplication rule
-Bot = ( s + s' ) and get uncertainty of 5.57mm using the uncertainty addition rule.

2) Then I divide the top by the bottom apply the uncertainty division rule.

Am I doing everything correctly?

2. Jan 15, 2012

### Simon Bridge

It can be smaller in absolute terms but I'd still expect it to be of the order of 1%.
You need to check the calculation ...

3. Jan 15, 2012

### Simon Bridge

You have a relation of the form $$\frac{1}{z} = \frac{1}{x} + \frac{1}{y} \Rightarrow z = \frac{xy}{x+y}$$... and you want to find $z \pm \sigma_z$ given $x \pm \sigma_x$ and $y \pm \sigma_y$

rule: proportional errors do not change with inverting the fractions

hence$$\frac{\sigma_{1/x}}{1/x} = \frac{\sigma_x}{x} \Rightarrow \sigma_{1/x} = \frac{\sigma_x}{x^2}$$... and the same for $\sigma_{1/y}$

rule: the variance of the sum is the sum of the variances
$$\Rightarrow \sigma_{1/z}^2 = { \sigma_{1/x}^2 + \sigma_{1/y}^2}$$
rule: proportional errors do not change with inverting the fractions
$$\frac{\sigma_z}{z} = \frac{\sigma_{1/z}}{1/z} \Rightarrow \sigma_z = z^2\sigma_{1/z}$$

... you should be able to take it from there.