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**Hint:**We know that Intensity of a wave is directly proportional to the square of amplitude. That is, $I \propto {a^2}$. where $I$ denotes the intensity of wave and $a$ denotes the amplitude.

In equation form it can be written as $I = c{a^2}$ where c is a constant. Since the ratio of intensity is given we can use this relation to find the ratio of amplitudes.

**Complete step by step answer:**

Let ${I_1}$ be the intensity of the first wave and ${I_1}$ be the intensity of the second wave.

Given, ${I_1}:{I_2} = 1:16$

Intensity of a wave is directly proportional to the square of amplitude.

$I \propto {a^2}$

That is,

$I = c{a^2}$ (1)

where c is a constant.

Let ${a_1}$ be the amplitude of the first wave. Then using equation(1) the intensity of the first wave ${I_1}$ can be written as ,

${I_1} = c{a_1}^2$ (2)

Let ${a_2}$ be the amplitude of the second wave Then using equation (1) intensity of second wave ${I_2}$ can be written as ,

${I_2} = c{a_2}^2$ (3)

Now let us divide equation (2) by (3). Then we get,

$\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{c{a_1}^2}}{{c{a_2}^2}}$

$\therefore \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{{a_1}^2}}{{{a_2}^2}}$

Now substitute the value of the ratio of intensities $\dfrac{{{I_1}}}{{{I_2}}}$ in the above equation. Then we get,

$\dfrac{1}{{16}} = \dfrac{{{a_1}^2}}{{{a_2}^2}}$

$ \Rightarrow \dfrac{{{a_1}}}{{{a_2}}} = \sqrt {\dfrac{1}{{16}}} $

$\therefore \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{1}{4}$

Therefore, the ratio of amplitudes of the waves is 1:4.

**So, the answer is option C**

**Note:**The equation for finding intensity is given as $I = 2{\pi ^2}\rho A{\upsilon ^2}{a^2}$where $\rho $ is the density of the medium, $A$ is the area, $\upsilon $ is the frequency, $a$ is the amplitude. While solving this question we assumed that $2{\pi ^2}\rho A{\upsilon ^2}$ is a constant. Since change in any of these factors is not mentioned it is okay to consider all those values as constant. But when they are changing, we cannot consider them as constants.

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