While the Beer's law equation is $A=a \cdot b \cdot c$, according to my book, $a$ is a coefficient which is:
a characteristic property of the analyte in a specific solvent, and its values will depend on the wavelength of absorption radiation.
This is quite unclear to me.
If we have for example $a= 1.5$, what does that exactly tell us ? What do we know when this coefficient is given ? Or is this just a constant that helps us to solve the equation?
To start with, I recommend to reformulate Beer-Lambert's law in a way more frequently met in (analytical) chemistry, i.e.
$\mbox{ABS} = c \cdot{} l \cdot{} \varepsilon = \log_{10} \frac{I_0}{I}$
because then it more obvious that the (dimensionless) absorption (ABS) recorded is the product of analyte concentration ($c$), optical path length ($l$) across the sample investigated, and the molecular extinction coefficient ($\varepsilon{}$). Absorption equally is defined as the logarithm (to base of ten) of the fraction of incoming light intensity ($I_0$) and light intensity that is transmitted across the sample ($I$).
Now to your question. While concentration $c$ and optical path length $l$ may be changed (the later, for example, by exchanging a measurement cell of 1 cm path length to one of 2 mm or 5 cm -- depending on the setup of the spectrometer), the molecular extinction coefficient $\varepsilon$ is indeed more valuable than just assistance to solve the equation. It relates to the ease the analyte may absorb irradiation.
UV-Vis spectroscopy investigates electronic transitions, prominently of $\pi{}$-systems (like of alkene double, and aromatic bonds), yet investigation of electronic transitions of $n$- (non-bonded, like of carbonyl-O electrons), $d$- (for example in transition metal complexes) and $\sigma{}$-electrons (in the short wavelength UV-range). In principle, these transitions are discrete, i.e. only photons of this attuned energy ($h\nu{}$) that fits the energy gap between the electronic ground state and the excited state are absorbed.
An analyte molecule may possess more than one allowed electronic transitions. They may occur simultaneously, with different efficiency. For example, many organic dyes have a strong absorption in the visible range, like indigo has one around 600 nm. Yet if you tune the wavelength of investigation to short wavelengths, below 250 nm, for example, absorption may still occur, too, yet to a lesser extent. (This eventually contributes to processes that indigo stained cloths fade over time, for example.)
Side note: The molecular extinction coefficient $\varepsilon$ depends on other, additional parameters than the "solvent" and "wavelength of investigation". Varying the pH-value, ionic strength of the solution, presence/absence of other dissolved molecules than of your analyte and solvent may alter $\varepsilon{}$, too. Do not forget, the concentration $c$ of your analyte may influence $\varepsilon{}$, too -- pointing to (concentration dependent) potential dimerisation / aggregation of your analyte (for example charge transfer complexes).
