Find how the intensity of light decreases as it passes

Find how the intensity of light decreases as it passes through a homogeneous transparent medium, given that the intensity lost in a thin layer is proportional to both the intensity entering that layer and the layer's thickness.

Step 1: Set up the differential equation

Let $I(x)$ be the intensity after traveling distance $x$ into the medium, and $\mu$ the (constant) absorption coefficient. The stated law is $-dI=\mu I\,dx$, i.e.

\[\frac{dI}{dx}=-\mu I.\]

Step 2: Separate and integrate

\[\frac{dI}{I}=-\mu\,dx\quad\Longrightarrow\quad \ln I=-\mu x+C.\]

Step 3: Apply the initial condition

At $x=0$, $I=I_0$, so $C=\ln I_0$, giving $\ln(I/I_0)=-\mu x$, i.e.

\[\boxed{I(x)=I_0e^{-\mu x}}\]

Step 4: Characteristic lengths

The half-thickness (depth at which intensity halves) solves $\tfrac12=e^{-\mu x_{1/2}}$:

\[x_{1/2}=\frac{\ln2}{\mu}\approx\frac{0.693}{\mu}.\]

The attenuation length (depth at which intensity falls to $1/e$) is simply $x_e=1/\mu$.

Step 5: Fraction absorbed and optical density

The fraction of light absorbed after distance $x$ is $1-e^{-\mu x}$. For a slab of thickness $L$, the transmission is $T=e^{-\mu L}$ and the optical density (absorbance) is $A=\log_{10}(1/T)=\mu L/\ln10\approx0.434\mu L$.