An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light = $$3 \times 10^{8}$$ m s$$^{-1}$$)

We begin by identifying the data given in the question. The emitted (source) frequency is $$f_0 = 10\ \text{GHz}$$ and the observer is moving directly toward the source with speed $$v = \tfrac12 c$$, where $$c = 3 \times 10^{8}\ \text{m s}^{-1}$$ is the speed of light.

Because the speed involved is a significant fraction of the speed of light, we must use the relativistic Doppler-shift formula for light. The formula for the frequency $$f'$$ measured by an observer moving with speed $$v$$ toward a stationary source is

$$f' = f_0 \sqrt{\frac{1 + \beta}{1 - \beta}}$$

where we have defined the dimensionless ratio

$$\beta = \frac{v}{c}.$$

Now we substitute the given speed into $$\beta$$:

$$\beta = \frac{v}{c} = \frac{\tfrac12 c}{c} = \frac12 = 0.5.$$

Next we compute the two expressions that appear in the square root:

$$1 + \beta = 1 + 0.5 = 1.5,$$ $$1 - \beta = 1 - 0.5 = 0.5.$$

We form their ratio:

$$\frac{1 + \beta}{1 - \beta} = \frac{1.5}{0.5} = 3.$$

Taking the square root gives

$$\sqrt{\frac{1 + \beta}{1 - \beta}} = \sqrt{3} \approx 1.732.$$

Finally we multiply this factor by the emitted frequency $$f_0$$ to obtain the frequency measured by the observer:

$$\begin{aligned} f' &= f_0 \times \sqrt{\frac{1 + \beta}{1 - \beta}} \\ &= 10\ \text{GHz} \times 1.732 \\ &= 17.32\ \text{GHz}. \end{aligned}$$

This value rounds to 17.3 GHz, which matches option D.

Hence, the correct answer is Option D.