= Solution
The <characteristic impedance> of a lossless <transmission line> is the ratio of the <voltage> to the current amplitude of a single forward-travelling wave. If $\mathcal L$ and $\mathcal C$ denote <inductance> and <capacitance per unit length>, the lossless telegrapher equations give
$$
\partial_z V=-\mathcal L\partial_t I,\qquad
\partial_z I=-\mathcal C\partial_t V,
\qquad \boxed{Z=\sqrt{\mathcal L/\mathcal C}},\qquad
v_p=(\mathcal L\mathcal C)^{-1/2}.
$$
A backward wave has $I_-=-V_-/Z$ when current is positive along $+z$. Thus <characteristic impedance> is not a resistor's <electrical resistance>: it describes travelling-wave amplitudes and determines <impedance matching>, power transfer and reflections. A semi-infinite line, or a line terminated by a matching load, presents that impedance at its input.
At a junction with no lumped branch, <voltage> and <electric current> are continuous:
$$
V_i+V_r=V_t,\qquad \frac{V_i-V_r}{Z_1}=\frac{V_t}{Z_2}.
$$
Eliminating $V_t$ gives the voltage <reflection coefficient>
$$
\boxed{r=\frac{V_r}{V_i}=\frac{Z_2-Z_1}{Z_2+Z_1}.}
$$
For real positive $Z_1$, incident and reflected waves occupy the same line, so their power ratio is $|r|^2$.
For the slightly mismatched <coaxial cables>, the change in <characteristic impedance> is
$$
\Delta Z=\frac{Z_{\rm vac}}{2\pi n}\ln\!\left(1+\frac{\Delta b}b\right)
\simeq\frac{Z_{\rm vac}}{2\pi n}\frac{\Delta b}b.
$$
Consequently
$$
r\simeq\frac{\Delta Z}{2Z}=\frac1{2\ln(b/a)}\frac{\Delta b}b,\qquad
\boxed{\mathcal R\simeq\frac{1}{4\ln^2(b/a)}\left(\frac{\Delta b}b\right)^2.}
$$
The requested power limit $0.01\%=10^{-4}$ implies $|\Delta b|/b<0.02\ln(b/a)$. Since $\ln(b/a)=2\pi nZ/Z_{\rm vac}=1.918$ for the stated line, \b[the adjacent cables' outer radii must match within approximately $3.84\%$]. If a tolerance is quoted separately as a symmetric manufacturing range $\pm\delta b$ and arbitrary worst-case pairs can differ by $2\delta b$, the corresponding individual tolerance is approximately \b[$\pm1.92\%$]. The leading calculation is for small mismatches; the exact logarithmic expression gives slightly asymmetric positive and negative limits. The vacuum <wave impedance> used here is approximately $376.73\ \Omega$.<Fundamental Physical Constants — Complete Listing by National Institute of Standards and Technology and others>[2022 CODATA adjustment, characteristic impedance of vacuum entry]{ref}
A needle is a shunt <electrical resistance> $R_n$, so the load seen at its location is the continuation of the line in parallel with $R_n$: $Z_L=ZR_n/(Z+R_n)$. The local voltage <reflection coefficient> and transmission amplitude are
$$
\boxed{r=-\frac{Z}{2R_n+Z}\simeq-\frac{Z}{2R_n}},\qquad t=1+r.
$$
For <periodic resistive loading of a transmission line>, the reflection from the next needle travels an additional round trip $2L$ and crosses each previous needle twice. Taking the first needle as phase reference, the <single-return approximation for periodic line reflections> therefore sums to
$$
R_{\rm first}=r\sum_{j=0}^\infty[t^2e^{i\phi}]^j
=\frac r{1-(1+r)^2e^{i\phi}},\qquad
\phi=2kL=\frac{4\pi\nu L}{v_p}.
$$
Here $v_p$ is the propagation <phase velocity> in the cable, $c/n$ for the stated nondispersive nonmagnetic filling. Squaring the amplitude gives the requested expression
$$
\boxed{\mathcal R_{\rm first}(\nu)=\left|\frac r{1-(1+r)^2e^{i\phi(\nu)}}\right|^2.}
$$
For a finite number $N$ of needles the amplitude has the additional numerator $1-[t^2e^{i\phi}]^N$. The displayed semi-infinite result assumes sufficiently many needles, no terminal reflection, and neglects intrinsic propagation loss.
For real $r<0$, write $q=t^2$. The denominator of the power coefficient is $1+q^2-2q\cos\phi$. It has constructive peaks at $\phi=2\pi j$ and minima at $\phi=(2j+1)\pi$, repeating with \b[frequency spacing $\Delta\nu=v_p/(2L)$]. Within this approximation,
$$
\mathcal R_{\max}=\frac1{(2+r)^2}\simeq\frac14,\qquad
\mathcal R_{\min}=\frac{r^2}{[1+(1+r)^2]^2}\simeq\frac{r^2}{4}.
$$
Weak individual shunts thus give narrow interference peaks, with phase width of order $|r|$.
The requested sum excludes paths that reflect repeatedly between different needles. It should not be read as an exact multiple-scattering result, particularly at the coherent peaks. For example the exact semi-infinite amplitude satisfies
$$
R=r+\frac{t^2e^{i\phi}R}{1-r e^{i\phi}R}.
$$
At $\phi=2\pi j$ this equation reduces to $(R+1)^2=0$: the exact ideal infinite array has unit power reflection there, rather than the first-return approximation's value near $1/4$. This does not change the requested geometric-series derivation, but identifies its physical limitation.
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