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**Hint**Loudness is nothing but a comparison of intensity on a logarithmic scale, while intensity is proportional to square of the amplitude of the wave (be it sound or be it light)

**Complete step-by-step solution**

When 2 sound waves interfere, they form a range of intensities because of the interference pattern. In this question we need to find the range of those values which is the difference in max and min volume. The maximum intensity is given by

\[I = {\text{ }}{I_1} + {I_2} + 2\sqrt {{I_1}{I_2}\cos \alpha } \]

So max intensity is obtained when \[\cos (\alpha ) = \max = 1\]

\[ \Rightarrow {I_{\max }} = {(\sqrt {{I_1}} + \sqrt {{I_2}} )^2}\]

As the 2 intensities are in ratio of 4:1, for some arbitrary constant of proportionality (\[{\text{x}}\]) let’s say:

\[{I_1} = 4x\] , \[{I_2} = x\]

\[{I_{\max }} = {(\sqrt {4x} + \sqrt x )^2}\]

\[{I_{\max }} = {(2\sqrt x + \sqrt x )^2}\]

\[{I_{\max }} = 9x\]

And the minimum intensity is given by when \[\cos (\alpha ) = \min = - 1\]

\[{I_{\min }} = {\text{ }}{I_1} + {I_2} - 2\sqrt {{I_1}{I_2}\cos \alpha } \]

\[

{I_{\min }} = {(\sqrt {{I_1}} - \sqrt {{I_2}} )^2} \\

\Rightarrow {I_{\min }} = {(\sqrt {4x} - \sqrt x )^2} \\

\Rightarrow {I_{\min }} = {(2\sqrt x - \sqrt x )^2} \\

\Rightarrow {I_{\min }} = x \\

\]

The difference is these 2 values will give us the correct option

\[{I_{\max }} - {I_{\min }} = {\text{ }}9x - x{\text{ }} = {\text{ }}8x\]

**Therefore the correct answer is option B.**

**Note**The intensity of light waves are related to amplitude by this relation,

$

\sqrt {\dfrac{{{I_1}}}{{{I_2}}}} = \dfrac{{{a_1}}}{{{a_2}}} = \left( {\dfrac{{\sqrt {\dfrac{{{I_{\max }}}}{{{I_{\min }}}}} + 1}}{{\sqrt {\dfrac{{{I_{\max }}}}{{{I_{\min }}}}} - 1}}} \right) \\

\dfrac{{{I_{\max }}}}{{{I_{\min }}}} = \left( {\dfrac{{{a_1} + {a_2}}}{{{a_1} - {a_2}}}} \right) \\

$

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