Numford
Error analysis

Error in the focal length of a lens

f = u v/(u+v). f is the focal length of a convex lens, u the object distance and v the image distance, both taken positive.

Error in a formula

Product, quotient, power. The percentage errors add. Each counts as many times as its power. Sum or difference. The absolute errors add. A percentage needs the value to become an absolute error. Both in one formula. Worked out with the values, as the sum of slope times error. This is the worst case. The answer is always the maximum error.
u
v

Maximum error in f

0.278

f = 20.00 ± 0.281.4%

  • u0.444 × 0.50.222
  • v0.111 × 0.50.0556
  • Totaladd them0.278

This mixes a sum with a product, so it is worked out with the values: the sum of |slope| times error, the worst case.

Worked example

f = 20.00 ± 0.28

Δf = |∂f/∂u| Δu + |∂f/∂v| Δv

This mixes a sum with a product, so it is worked out with the values: the sum of |slope| times error, the worst case. The lens formula 1/f = 1/u + 1/v has a sum in it, so the error is worked out with the values.

Related

Other formulas

Measured the same thing several times? Use repeated readings, or find the least count of your instrument.

Questions

Error in the focal length of a lens: quick answers

What is the formula for the error in the focal length of a lens?

f is the focal length of a convex lens, u the object distance and v the image distance, both taken positive. With f = u v/(u+v), the maximum error is given by Δf = |∂f/∂u| Δu + |∂f/∂v| Δv. This mixes a sum with a product, so it is worked out with the values: the sum of |slope| times error, the worst case.

How do I work out the error with real numbers?

Take u = 30 ± 0.5 and v = 60 ± 0.5. The maximum error in f is 0.222 + 0.0556 = 0.278, so f = 20.00 ± 0.28.

Why are the errors added and never subtracted?

The error shown is the largest possible one. In the worst case every measurement is off in the direction that pushes the result the same way, so the errors pile up. They are added even when the quantity is divided or subtracted.

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