Given that quality factor Q = 100
The intermediate frequency fi = 455KHz
Incoming frequency fs = 1MHz
The image frequency fsi = fs + 2fi = 1 × 106 + 2 × 455 × 103 = 1910KHz
Again we know that the relation
\[ \rho = \frac{f_{si}}{f_s}- \frac{f_{s}}{f_{si}}=\frac{1910}{1000}-\frac{1000}{1910}=1.910-0.524=1.386 \]
The image frequency rejection ratio
\[ \alpha=\sqrt{1+Q^2\rho^2}=\sqrt{1+100^{2} \times1.386^{2}}=138.6 \]