A bothersome feature of many physical measurements is the presence of a background signal (commonly called "noise"). In Part 2.2.4 of the experiment, some light that reflects off the apparatus or from neighboring stations strikes the photometer even when the direct beam is blocked. In addition, due to electronic drifts, the photometer does not generally read 0.0 mV even in a dark room. It is necessary, therefore, to subtract off this background level from the data to obtain a valid measurement. Suppose the measured background level is 4.9 mV. A signal of 20.3 mV is measured at a distance of 29.5 mm and 17 mV is measured at 33 mm. Correct the data for background and normalize the data to the maximum value. What is the normalized corrected value at 33 mm?
The background-corrected light measurement at 29.5 mm is 20.3 - 4.9 = 15.4 mV
The corrected light measurement at 33 mm is 17.0 - 4.9 = 12.1 mV
You need the maximum signal to normalize the data. You have not provided that data.
12.1/15.4= .785
^from drwls's answer above
To correct the data for background, you need to subtract the measured background level from the measured signal. In this case, the background level is 4.9 mV.
For the measurement at 33 mm, the measured signal is 17 mV. To correct for background, subtract the background level from the measured signal:
Corrected signal at 33 mm = 17 mV - 4.9 mV = 12.1 mV
To normalize the data to the maximum value, you need to divide each corrected signal by the maximum signal value among all the measurements. In this case, the maximum measured signal is 20.3 mV.
Normalized corrected value at 33 mm = (Corrected signal at 33 mm) / (Maximum measured signal) = 12.1 mV / 20.3 mV = 0.596 = 59.6% (rounded to the nearest decimal place)
Therefore, the normalized corrected value at 33 mm is approximately 59.6%.