A) \[616\,\overset{o}{\mathop{A}}\,\]
B) \[616\,\overset{o}{\mathop{A}}\,\]
C) \[5500\,\overset{o}{\mathop{A}}\,\]
D) \[6160\,\overset{o}{\mathop{A}}\,\]
Correct Answer: B
Solution :
The Dopplers shift (sometimes called the Dopplers effect) is a change in frequency of the emitted waves produced by motion of an emitting source relative to an observer. The perceived frequency (n) is given by \[n=n\left( \frac{v+c}{c} \right)\] where n is source frequency and v is velocity. \[\therefore \] \[\frac{n}{n}=\frac{c+v}{c}\] Also frequency (n) \[=\frac{c}{wavelength\text{ }\left( \lambda \right)}\] \[\therefore \] \[\frac{\lambda }{\lambda }=\frac{c}{c+v}\] \[\Rightarrow \] \[\lambda =\frac{\lambda \,c}{c+v}\] Given, \[\lambda =5500\,\overset{o}{\mathop{A}}\,,\,\,v=0.5\,c\] \[\therefore \] \[\lambda =\frac{5500\,\,c}{c+0.5}=3667\,\overset{o}{\mathop{A}}\,\] Hence, \[\Delta \,\lambda =5500-3667=1833\,\overset{o}{\mathop{A}}\,\] Note: When frequency increases, wavelength decreases.
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