The physical sizes of the transmitter and receiver antenna in a communication system are:

We recall the basic fact from electromagnetic wave theory that an antenna radiates most efficiently when its physical length is a substantial fraction of the signal’s wavelength, usually a half-wave $$\left(\dfrac{\lambda}{2}\right)$$ or a quarter-wave $$\left(\dfrac{\lambda}{4}\right).$$

First, we write the relation that connects wavelength and frequency. For any electromagnetic wave travelling in free space we have the universal formula

$$\lambda \, f \;=\; c,$$

where $$\lambda$$ is the wavelength, $$f$$ is the frequency of that wave, and $$c = 3.0 \times 10^{8}\,\text{m s}^{-1}$$ is the speed of light.

Re-arranging the above equation to express the wavelength we obtain

$$\lambda \;=\; \dfrac{c}{f}.$$

Now an efficient transmitting or receiving antenna is usually made one-quarter of this wavelength. Stating that design rule explicitly,

$$\text{Antenna length} \; l \;=\; \dfrac{\lambda}{4}.$$

Substituting the value of $$\lambda$$ from the earlier relation, we get

$$l \;=\; \dfrac{1}{4}\,\left(\dfrac{c}{f}\right) \;=\; \dfrac{c}{4f}.$$

Here the frequency $$f$$ referred to is the carrier frequency, because it is the carrier wave that the antenna actually radiates or receives. The modulation frequency merely superimposes information onto the carrier; it does not determine the dimensions of the antenna.

From the expression $$l = \dfrac{c}{4f}$$ we clearly see that the antenna length $$l$$ is inversely proportional to the carrier frequency $$f$$:

$$l \;\propto\; \dfrac{1}{f}.$$

Therefore, as the carrier frequency increases, the required physical size of both the transmitter and receiver antennas decreases, and vice versa.

This observation matches exactly with Option D, which states that the physical sizes are inversely proportional to the carrier frequency.

Hence, the correct answer is Option D.