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**Hint:**If the source of the wave is moving with respect to the observer then there is a change in the frequency. This can be determined using Doppler’s effect. Doppler’s effect is defined as the change in the frequency of the waves if the source and the observer are moving with respect to each other. The Doppler’s effect can be observed in any kind of waves like sound or light etc.

**Complete step by step solution:**

If the observer and the object is moving, then as per Doppler’s effect, the frequency of the microwave is given by the formula:

$ \Rightarrow f = {f_0}\sqrt {\dfrac{{c + v}}{{c - v}}} $---(i)

Where $f$is the change in frequency

${f_0}$is the original frequency

$c$is the speed of the light

$v$is the velocity of the observer

It is given that the speed of the observer is half the speed of light. If the speed of the microwave is ‘c’, then the speed of the observer will be

\[ \Rightarrow v = \dfrac{c}{2}\]

Also given that the original frequency of the waves is ${f_0} = 10GHz$

Substituting the given values in equation (i),

$ \Rightarrow f = 10\sqrt {\dfrac{{c + \dfrac{c}{2}}}{{c - \dfrac{c}{2}}}} $

$\Rightarrow f = 10\sqrt {\dfrac{{\dfrac{{2c + c}}{2}}}{{\dfrac{{2c - c}}{2}}}} $

$\Rightarrow f = 10\sqrt {\dfrac{{\dfrac{{3c}}{2}}}{{\dfrac{c}{2}}}} $

\[\Rightarrow f = 10\sqrt {\dfrac{{3c}}{2} \times \dfrac{2}{c}} \]

$\Rightarrow f = 10\sqrt 3 $

$\Rightarrow f = 10 \times 1.73$

$\Rightarrow f = 17.3GHz$

The frequency of the microwave measured by the observer will be $ = 17.3GHz$.

**Option A is the right answer.**

**Note:**It is important to note that in case of Doppler’s effect the actual frequency of the waves does not change. When a light source emits the light and the waves move towards the observer then the waves seem compressed to the observer. On the other hand, when the light waves move away from the observer then they get stretched out.

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