A) The frequency of the reflected wave is \[\frac{\nu (c+v)}{c-v}\]
B) The wavelength of the reflected wave is \[\frac{c(c-v)}{\nu (c+v)}\]
C) The number of waves striking the surface per second is \[\frac{\nu (c+v)}{c}\]
D) The number of beats heard by a stationary listener to the left of the reflecting surface is \[\frac{\nu \ v}{c-v}\]
Correct Answer: B
Solution :
Number of waves striking the surface per second (or the frequency of the waves reaching surface of the moving target )\[n'=\frac{(c+v)}{\lambda }\]\[=\frac{\nu (c+v)}{c}\] Now these waves are reflected by the moving target (Which now act as a source). Therefore apparent frequency of reflected second \[n''=\left( \frac{c}{c-v} \right)n'\]\[=\nu \left( \frac{c+v}{c-v} \right)\] The wavelength of reflected wave \[=\frac{c}{n''}=\frac{c(c-v)}{\nu (c+v)}\] The number of beats heard by stationary listener \[=n''-\nu =\nu \left( \frac{c+v}{c-v} \right)-\nu =\frac{2\nu v}{(c-v)}\]
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