Error propagation for value not directly measurable

[sunrah]

Homework Statement

This should be very simple:
Given the following (boundary frequency for photoelectric effect):

\nu = \frac{\phi}{h}

what would be the error on \nu?

Homework Equations

The Attempt at a Solution

\varphi and h are both determined through linear regression (y = mx + c). Where

h =em and \varphi = -ec. The errors on m and c are supplied courtesy of computer software.

My understanding is that the product/division rule for error propagation can be used:

\delta \nu = \nu\sqrt{(\frac{\delta \varphi}{\varphi})^{2}+(\frac{\delta h}{h})^{2}}

my instructor disagrees

 

[vela]

Sounds right to me. What's your instructor's idea for how to propagate the error?

 

[sunrah]

well, i spoke to her now and she says that the systematic error on the voltage (digital multimeter) should be included.

eU = h\nu + \varphi (1)

where \varphiis the work function.
The boundary frequency is given as \nu_{b} = \frac{\varphi}{h} (2)

Of course we can write the \nu_{b} in terms of voltage using (1), but we determined \varphi and h using linear regression which also supplied the standard error on those values. I'm not really sure how to determine the errors on \nu_{b} now. Should I ignore the error on \varphi, combine (1) and (2) and just use the error for the voltage provided by the multimeter manufacturer?

\delta \nu_{b} = \nu_{b} \sqrt{(\frac{\delta h}{h})^2 + (\frac{\delta U}{U})^2}