(a) As given that intensity of a light pulse $\mathrm{I}=\mathrm{I}_0 \mathrm{e}^{-\mathrm{ax}}$ where, $\mathrm{I}_0$ is the intensity at $x=0$ and $\alpha$ is constant. According to the question remaining Intensity reduced by $75 \%$ so, remaining
$$
\mathrm{I}=25 \% \text { of } \mathrm{I}_0=\frac{25}{100} \mathrm{I}_0=\frac{\mathrm{I}_0}{4} \text { So, }\left(\mathrm{I}=\frac{\mathrm{I}_0}{4}\right)
$$
Using the formula mentioned in the question,
$$
\begin{aligned}
&\mathrm{I}=\mathrm{I}_0 \mathrm{e}^{-\alpha \mathrm{x}} \\
&\frac{\mathrm{I}_0}{4}=\mathrm{I}_0 \mathrm{e}^{-\alpha \mathrm{x}}
\end{aligned}
$$
So, $\frac{1}{4}=\mathrm{e}^{-\alpha \mathrm{x}}$
Taking $\log$ on both sides, we get
$$
\begin{array}{ll}
(\ln 1-\ln 4)=\log _e e^{-a x} & \\
\ln 1-\ln 4=-\alpha x \ln e & \left(\because \ln { }_e=1\right) \\
0-\ln 4=-\alpha x & (\because \ln 1=0) \\
x=\frac{\ln 4}{\alpha} &
\end{array}
$$
So, at distance $\mathrm{x}=\frac{\ln 4}{\alpha}$, the intensity is reduced by $75 \%$ of initial intensity.
(b) Let $\alpha$ be the attenuation in $\mathrm{dB} / \mathrm{km}$. If $\mathrm{x}$ is distance travelled by signal,
Then, $10 \log _{10}\left(\frac{\mathrm{I}}{\mathrm{I}_0}\right)=-\alpha \mathrm{x}$
where, $\mathrm{I}_0$ is the intensity initially.
According to the question Intensity reduced by $50 \%$ so remain
$\mathrm{I}=\left(50 \%\right.$ of $\left.\mathrm{I}_0\right)=\frac{\mathrm{I}_0}{2}$ and $\mathrm{x}=50 \mathrm{~km}$
Putting the value of $x$ in Eq. (i), we get
$$
\begin{aligned}
&10 \log _{10}\left(\frac{\mathrm{I}_0}{2 \mathrm{I}_0}\right)=-\alpha \times 50 \\
&10[\log 1-\log 2]=-50 \alpha
\end{aligned}
$$

$$
\frac{10 \times 0.3010}{50}=\alpha, \alpha=0.0602 \mathrm{~dB} \text { per km }
$$
So, The attenuation of signal for an optical fibre channel. $\alpha=0.0602 \mathrm{~dB} / \mathrm{km}$