(a) The true absorbance of a sample is 1.500, but $\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{\%}$stray light reaches the detector. Find the apparent transmittance and apparent absorbance of the sample. (b) How much stray light can be tolerated if the absorbance error is not to exceedat a true absorbance of? (c) A research-quality spectrophotometer has a stray light level of $\mathbf{<}\mathbf{0}\mathbf{.}\mathbf{00005}\mathbf{\%}$,at. What will be the maximum absorbance error for a sample with a true absorbance of 2? Of 3?

Absorbance is defined as "the logarithm of the ratio of incident to transmitted radiant power through a sample". Alternatively, for samples which scatter light, absorbance may be defined as "the negative logarithm of one minus absorptance , as measured on a uniform sample".

The apparent transmittance is:$\mathrm{T}=\frac{\mathrm{T}+\mathrm{S}}{{\mathrm{T}}_0+\mathrm{S}}=\frac{0.0316+0.05}{1+0.005}=0.0364$

The apparent absorbance is :$\mathrm{A}=-\log (\mathrm{T})=-\log (0.0364)=1.439$

The apparent transmittance is: $\mathrm{T}={10}^{-\mathrm{A}}={10}^{-1.999}=0.01002305$

$\mathrm{T}=\frac{\mathrm{T}+\mathrm{S}}{{\mathrm{T}}_0+\mathrm{S}}$

The S is : $\frac{0.010+\mathrm{S}}{1+\mathrm{S}}=0.01002305$

$0.010+\mathrm{S}=0.01002305+0.01002305$.S

$\mathrm{S}=2.328\cdot {10}^{-5}$

$\mathrm{S}=2.38\cdot {10}^{-3\%}\mathrm{\%}$

$$\begin{gathered} \mathrm{T}=\frac{\mathrm{T}+\mathrm{S}}{{\mathrm{T}}_0+\mathrm{S}} \\ \mathrm{T}=\frac{0.010+0.0000005}{1+0.0000005} \\ \mathrm{T}=0.010000495 \end{gathered}$$

The apparent absorbance error is: $\mathrm{A}=-\log (\mathrm{T})=-\log (0.010000495)=1.999978$

So, the absorbance error is:$2-1.999978=2.2\cdot {10}^{-5}$

If the absorbance is 3 ,the apparent transmittance is:

$$\begin{gathered} \mathrm{T}=\frac{\mathrm{T}+\mathrm{S}}{{\mathrm{T}}_0+\mathrm{S}} \\ \mathrm{T}=\frac{0.001+0.0000005}{1+0.0000005} \\ \mathrm{T}=0.001000495 \end{gathered}$$

The apparent absorbance is: $\mathrm{A}=-\log (\mathrm{T})=-\log (0.001000495)=2.999785$

So, the absorbance error is: $3-2.999785=2.15\cdot {10}^{-4}$