- #1

ChiralSuperfields

- 1,281

- 136

- Homework Statement
- Please see below

- Relevant Equations
- ##NA = \bar {NA} ± \Delta NA##

For this data,

I am trying to find the overall absolute uncertainty of NA, where NA is the numerical aperture: ##\tan \theta_{NA} = \frac{R}{L}## and ##NA = \sin \theta_{NA}##

Since ##NA = \bar {NA} ± \Delta NA ## (Where ##\bar {NA}## is the best estimate for NA and ##\Delta NA## is the uncertainty in NA) we take the absolute values of the values for the error in NA shown in bold? This is because if we solve for the absolute uncertainty,

##NA - \bar {NA} = ±\Delta NA##

Then take the absolute value of both sides,

## |NA - \bar {NA} | = \Delta NA##

Any help much appreciated!

Many thanks!

I am trying to find the overall absolute uncertainty of NA, where NA is the numerical aperture: ##\tan \theta_{NA} = \frac{R}{L}## and ##NA = \sin \theta_{NA}##

Case | R(cm) | L(cm) | R/L | theta_NA[rad] | NA | error in NA |

R0,L0 | 0.5 | 0.5 | 1 | 0.785398163 | 0.707106781 | 0 |

Rmax,Lmax | 0.7 | 0.55 | 1.272727273 | 0.904827089 | 0.786318339 | 0.079211558 |

Rmax,Lmin | 0.7 | 0.45 | 1.555555556 | 0.999458847 | 0.841178475 | 0.134071694 |

Rmin,Lmax | 0.3 | 0.55 | 0.545454545 | 0.499346722 | 0.478852131 | -0.228254651 |

Rmin,Lmin | 0.3 | 0.45 | 0.666666667 | 0.588002604 | 0.554700196 | -0.152406585 |

Since ##NA = \bar {NA} ± \Delta NA ## (Where ##\bar {NA}## is the best estimate for NA and ##\Delta NA## is the uncertainty in NA) we take the absolute values of the values for the error in NA shown in bold? This is because if we solve for the absolute uncertainty,

##NA - \bar {NA} = ±\Delta NA##

Then take the absolute value of both sides,

## |NA - \bar {NA} | = \Delta NA##

Any help much appreciated!

Many thanks!