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An absorption measurement with a 1 cm path length yields a reading of 0.002 absorbance units with a noise of 0.0005 absorbance units and a mean noise reading of 0 for 5 scans averaged together. Assuming the noise for a single scan does not change if the path length is increased to 5 cm and the number of scans is increase to 65, what is the signal to noise of the new measurement. Remember signal averaging

User Chrisber
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The signal-to-noise ratio (S/N) is a measure of the strength of the signal relative to the noise. In this case, the signal is the absorbance reading of 0.002 absorbance units, and the noise is the standard deviation of the noise, which is 0.0005 absorbance units.

The S/N for the 1 cm path length measurement is:

S/N = (signal) / (noise) = 0.002 absorbance units / 0.0005 absorbance units = 4

If the path length is increased to 5 cm and the number of scans is increased to 65, the signal will also increase by a factor of 5, but the noise will not change. Therefore, the signal will be 5 times larger and the noise will be the same, so the S/N will increase.

The signal for the new measurement is 5 * 0.002 = 0.01 absorbance units

The S/N for the new measurement is:

S/N = (signal) / (noise) = 0.01 absorbance units / 0.0005 absorbance units = 20

When you increase the path length by 5 times, the signal increases by 5 times, but the noise remains the same. Therefore, the signal-to-noise ratio will increase by a factor of 5. Similarly when you increase the number of scans, you are averaging the signal, which will reduce the noise, but the signal will remain the same. So the signal-to-noise ratio will increase.

User NTyler
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