Answer
77.7k+ views
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) \\
$
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) \\
$
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write an article on the need and importance of sports class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Choose the one which best expresses the meaning of class 9 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)