45tt. I do not like your chances.
We do not know the output power so we will start with some assumptions. The transmitter operates continuously and is the base station for a fleet of mobile vehicle radios. It is most unlikely to radiate more than about 30 watts, as range at that frequency is decided by line-of-sight. The transmit antenna will probably be an omni-directional whip and may be optimised to transmit towards the horizon rather than overhead, but then the power will be reduced as the regulations will specify “effective isotropically radiated power” (EIRP), not the total power. You will be close to the horizon for a 2 metre transmitter that is one mile away.
Now you want to catch (2watt / 30watt) = one fifteenth = 6.67% of the effective power transmitted.
You will therefore need to cover one 15th of the transmitters horizon. At a distance of one mile that will be 1mile * TwoPi / 15 = 0.4186 mile long, and quite high. Since this is much greater than the 2 metre wavelength it will need to take the form of an elliptical wire fence reflector screen with vertical wires spaced approximately every fifth of a wavelength (15 inches). The ellipse should be drawn on the map with the transmit site at one focus and your receive site at the other. Only part of that fence need be built, namely the part behind you when you face the transmitter, but it will need to have an opening of at least half a mile wide facing the transmitter. It's position will need to be accurate to about one tenth of a wavelength (8 inches). That might deliver a maximum of 2 watt to your whip antenna.
You could probably do better by tuning to a longer wavelength, higher power (10kW) broadcast transmitter, even though it is further away. It would also operate continuously.
Remember that there is a nuclear power source that transmits in the optical spectrum. A 10 watt solar panel, (with a small lead acid storage battery to get you through the night), would be a much more economic solution.