0
$\begingroup$

I need to find the focal length of a lens by using equation 1/u + 1/v=1/f I have: u= 50+-3 mm v= 200+-5 mm I calculate the value of f as 40mm. Now i need to find the uncertainty in this value. I have two approaches, but only the second one is correct. I do not know what is wrong with the first one.

FIRST APPROACH : since f=(uv)/(u+v) Delta f/f= Fractional error of f= fractional error of u+ fractional error of v + fractional error of (u+v)

From this the uncertainty is 4.7 mm

SECOND APPROACH:we have Fractional error of 1/f = fractional error of f So delta( 1/f) = delta(f)/f^2 (*)

Similarly (*) is true for u and v in place of f

We have : delta(1/f) = delta(1/u) + delta(1/v)

So delta(f)/f^2= delta(u)/u^2 + delta(v)/v^2

From this delta(f) is 2.1mm which is correct

What is wrong with my first attempt?

$\endgroup$
0

1 Answer 1

2
$\begingroup$

The problem with your first approach is that you are assuming that the uncertainties in $u$, $v$ and $u+v$ are independent, when clearly they are not, they are highly positively correlated (when they are all positive). Hence you overestimate the uncertainty.

I should just add that I think both of your approaches are incorrect if you understand the error bar to mean the standard deviation of your estimate. Independent uncertainties should be combined in quadrature. I get $\delta F= 1.9$ mm.

$\endgroup$
3
  • $\begingroup$ How can I know that u,v and u+v are not independent. Why can I use the first approach in case w=sqrt(g/l)? Thank $\endgroup$ Commented Jan 2, 2016 at 10:18
  • 2
    $\begingroup$ Because $u+v$ depends on the values of $u$ and $v$ !? In your second example, presumably $g$ and $l$ are independent variables. $\endgroup$
    – ProfRob
    Commented Jan 2, 2016 at 12:18
  • $\begingroup$ @trunghiếulê how you have written this 'we have Fractional error of 1/f = fractional error of f So delta( 1/f) = delta(f)/f^2 (*)' $\endgroup$
    – Koolman
    Commented Dec 3, 2016 at 12:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.