A) \[{{\lambda }_{L}}={{\lambda }_{S}}\]
B) \[{{\lambda }_{L}}>{{\lambda }_{S}}\]
C) \[{{\lambda }_{L}}<{{\lambda }_{S}}\]
D) \[{{\lambda }_{L}}=2{{\lambda }_{S}}\]
Correct Answer: B
Solution :
Frequency of light wave = f Frequency of sound wave = f \[\therefore \]Wavelength of light waves \[{{\lambda }_{L}}=c/f\] ...(i) where, c = velocity of light and wavelength of sound waves \[{{\lambda }_{S}}=v/f\] v = velocity of sound \[\therefore \] \[\frac{{{\lambda }_{L}}}{{{\lambda }_{S}}}=\frac{c/f}{v/f}=\frac{c}{v}\] but \[c>>>v\Rightarrow \frac{c}{v}>>>1\] \[\Rightarrow \] \[\frac{{{\lambda }_{L}}}{{{\lambda }_{S}}}>>1\Rightarrow {{\lambda }_{L}}>>{{\lambda }_{S}}\]You need to login to perform this action.
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