In spectrophotometry, which expression correctly relates absorbance to transmittance when %T is used?

In spectrophotometry, which expression correctly relates absorbance to transmittance when %T is used?

• A. A = abc
• B. (Au/Cu) = (As/Cs)
• C. A = 2 - log %T
• D. A = log %T

Answer: (C)

Absorbance grows as transmittance drops, and its relation to light transmission is A = -log10(T), where T is the fraction of light that passes through (0 < T ≤ 1). If you’re using percent transmittance, %T, then T = %T/100. Substituting gives A = -log10(%T/100) = -log10(%T) + log10(100) = 2 − log10(%T). So the correct expression is A = 2 − log10(%T). For example, if %T = 50, then A ≈ 2 − log10(50) ≈ 0.301, which matches A = -log10(0.5). The other forms fail because they omit the negative sign, or they don’t account for dividing by 100 inside the logarithm.

Absorbance grows as transmittance drops, and its relation to light transmission is A = -log10(T), where T is the fraction of light that passes through (0 < T ≤ 1). If you’re using percent transmittance, %T, then T = %T/100. Substituting gives A = -log10(%T/100) = -log10(%T) + log10(100) = 2 − log10(%T). So the correct expression is A = 2 − log10(%T).

For example, if %T = 50, then A ≈ 2 − log10(50) ≈ 0.301, which matches A = -log10(0.5). The other forms fail because they omit the negative sign, or they don’t account for dividing by 100 inside the logarithm.