- #1

Terry Bing

- 48

- 6

- Homework Statement
- The focal length of a spherical mirror is given by ##\frac{1}{f}=\frac{1}{v}+\frac{1}{u}## . If ##u=30\pm 0.3## cm and ##v=60\pm 0.6## cm. Find the percentage error in the measurement of focal length of the mirror .

- Relevant Equations
- We estimate errors by using formulae for differentials.

If I write ##f=\frac{uv}{u+v}## and then take differentials on both sides, I get ##\frac{df}{f}=\frac{du}{u}+\frac{dv}{v}+\frac{du+dv}{u+v}##, I get the fractional error as 0.03. (I have replaced the negative signs that come as a result of quotient rule with positive signs, since we are asked to estimate the

If instead I write ##\frac{1}{f}=\frac{1}{v}+\frac{1}{u}## and then take differentials on both sides, then I get ##\frac{df}{f^2}=\frac{dv}{v^2}+\frac{du}{u^2}##. Using this, I get the fractional error to be 0.01. Which of these is a better estimate of the error? Also, why are there two different answers based on how I write the function. Basically I am taking first order approximations in both cases, right?

*maximum*error. This leads to the formula that when measurements are multiplied or divided together, their relative errors just add up.)If instead I write ##\frac{1}{f}=\frac{1}{v}+\frac{1}{u}## and then take differentials on both sides, then I get ##\frac{df}{f^2}=\frac{dv}{v^2}+\frac{du}{u^2}##. Using this, I get the fractional error to be 0.01. Which of these is a better estimate of the error? Also, why are there two different answers based on how I write the function. Basically I am taking first order approximations in both cases, right?

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