The minimum receivable power of the receiver is $2×10^{-13}$ Watt. What is the smallest diameter of the antenna reflector could have, assuming it to be full parabolid with η=0.65.

Given

f=1.5 GHz

Pt = 2.5 MW

$R_{max}=100nmi=100×1.8518=185.18 km \\ σ=1m^2 \\ P_r=2×10^{-13} Watt \\ d=? \\ η=0.65$

we know that,

$R_{max}=\bigg[\frac{pt Ae^2 σ}{4π λ^2 δ_{min}}\bigg] \\ 185.18×10^3= \frac{(2.5×10^6×Ae^2×1)}{4π×\bigg(\frac{3×10^8}{1.5×10^9}\bigg)^2×2×10^{-13}} \\ ∴Ae^2= \frac{(1.85×10^5 )^4×4π×(40×10^{-3})×2×10^{-13}}{2.5×10^6 } \\ Ae=6.8765$

But $Ae=η.A$

$η.A=6.8765 \\ 0.65×πr^2=6.8765 \\ ∴r^2=\frac{6.8765}{π×0.65} \\ r^2=3.3674 \\ ∴r=1.8350m. \\ ∴d=2r=3.670m \\ d=3.670 m$