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Question

Physics Question on Youngs double slit experiment

The ratio of intensities between two coherent sound sources is 4:14 : 1. The difference of loudness in decibels between maximum and minimum intensities, when they interfere in space, is

A

10log210 \log 2

B

20log320 \log 3

C

10log310 \log 3

D

20log220 \log 2

Answer

20log320 \log 3

Explanation

Solution

Given I1I2=41 \frac{I_{1}}{I_{2}}=\frac{4}{1} We know Ia2I \propto a^{2} a12a22=I1I2=41\therefore \frac{a_{1}^{2}}{a_{2}^{2}} =\frac{I_{1}}{I_{2}}=\frac{4}{1} a1a2=21\frac{a_{1}}{a_{2}}=\frac{2}{1} ImaxImin=(a1+a2)2(a1a2)2\therefore \frac{I_{\max }}{I_{\min }} =\frac{\left(a_{1}+a_{2}\right)^{2}}{\left(a_{1}-a_{2}\right)^{2}} =(2+121)2=\left(\frac{2+1}{2-1}\right)^{2} =(31)2=91=\left(\frac{3}{1}\right)^{2}=\frac{9}{1} Therefore, difference of loudness is given by L1L2=10logImaxImin=10log(9)L_{1}-L_{2} =10 \log \frac{I_{\max }}{I_{\min }}=10 \log (9) =10log32=20log3=10 \log 3^{2}=20 \log 3