= Solution
An <ideal operational amplifier> has infinite differential open-loop gain, infinite <input impedance>, zero <output impedance> and unlimited <bandwidth> and output capability, and zero <input offset voltage> and <input bias current>. In a stable negative-feedback circuit operating linearly, its input currents vanish and $V_-=V_+$. This equality is a feedback consequence, not a rule for a saturated or open-loop amplifier. Here the grounded non-inverting input makes the inverting node a virtual earth.
The <inverting two-pole band-pass filter> has a series <resistor> and <capacitor> input, while the feedback arm is a parallel <resistor> and <capacitor>. With Laplace variable $s$,
$$
Z_{\rm in}=R_1+\frac1{sC_1},\qquad
Z_f=\frac{R_2}{1+sR_2C_2},\qquad
\boxed{H(s)=-\frac{Z_f}{Z_{\rm in}}=-K\frac{s\tau_1}{(1+s\tau_1)(1+s\tau_2)}},\quad K=R_2/R_1.
$$
<Kirchhoff's current law> at virtual earth gives the minus sign. Substituting $s=i\omega$ yields
$$
|H|^2=K^2\frac{\omega^2\tau_1^2}{(1-\omega^2\tau_1\tau_2)^2+\omega^2(\tau_1+\tau_2)^2}.
$$
It vanishes at zero and infinite <frequency>, with low-frequency $|H|^2\simeq(\omega R_2C_1)^2$ and high-frequency $|H|^2\simeq1/(\omega R_1C_2)^2$. Differentiation gives the single maximum
$$
\boxed{\omega_0=\frac1{\sqrt{\tau_1\tau_2}},\quad
|H|_{\max}=K\frac{\tau_1}{\tau_1+\tau_2}}.
$$
For $\tau_1\gg\tau_2$, the broad interval $1/\tau_1\ll\omega\ll1/\tau_2$ has $|H|\simeq R_2/R_1$. For $\tau_2\gg\tau_1$, the interval $1/\tau_2\ll\omega\ll1/\tau_1$ instead has $|H|\simeq C_1/C_2$. The same rising and falling asymptotes apply outside either plateau.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2013/paper-1-band-pass.png]
{title=Original ideal filter power gains for different time-constant ratios, showing the finite selectivity limit}
Writing $a=\tau_1+\tau_2$ and $b=\tau_1\tau_2$, the half-power <frequencies> and <quality factor> are
$$
\omega_\pm=\frac{\sqrt{a^2+4b}\pm a}{2b},\quad
\Delta\omega=\omega_+-\omega_-=\frac ab,\quad
\boxed{Q=\frac{\omega_0}{\Delta\omega}=\frac{\sqrt{\tau_1\tau_2}}{\tau_1+\tau_2}\le\frac12}.
$$
Equal time constants give the narrowest fractional pass-band achievable here, with $\omega_\pm=(\sqrt2\pm1)\omega_0$ and
$$
\boxed{|H|_{\max}=\frac{R_2}{2R_1}=\frac{C_1}{2C_2}\quad(\tau_1=\tau_2)}.
$$
\b[This topology cannot make a genuinely sharp, high-$Q$ band-pass.] Its <poles> are real, and the <selectivity bound for a two-real-pole band-pass> prevents arbitrarily narrow selection. A more selective resonant filter needs additional circuit structure. The gain above answers the component-ratio question for the sharpest available setting without assuming an impossible limit.
Real amplifiers have finite frequency-dependent gain and phase lag, so virtual earth becomes inaccurate, the response changes and stability must be checked. Gain-bandwidth limits, <slew rate> for large <signals>, output loading and <amplifier saturation> also matter. Component tolerances move the corners; <capacitor> parasitics, stray <capacitance>, <resistor> <noise> and input <noise> become significant. The ideal high-frequency tail should not be extrapolated beyond the amplifier and lumped-component operating ranges.
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