
The ratio of intensities between two coherent sound sources is 4:1.The difference of loudness in dB between maximum and minimum intensities when they interfere in space is
(A) 10 log(2)
(B) 20 log(3)
(C) 10 log(3)
(D) 20 log(2)
Answer
133.2k+ 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
Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main Books 2023-24: Best JEE Main Books for Physics, Chemistry and Maths

JEE Main 2023 April 13 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 11 Shift 2 Question Paper with Answer Key

JEE Main 2023 April 10 Shift 2 Question Paper with Answer Key

JEE Main 2023 (April 11th Shift 2) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Current Loop as Magnetic Dipole and Its Derivation for JEE

Inertial and Non-Inertial Frame of Reference - JEE Important Topic

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

Motion In A Plane: Line Class 11 Notes: CBSE Physics Chapter 3

Waves Class 11 Notes: CBSE Physics Chapter 14
