\(n' = \frac{{(c + v)}}{\lambda }\)\( = \frac{{\nu (c + v)}}{c}\)
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 \(n = \left( {\frac{c}{{c - v}}} \right)n'\)
The number of beats heard by stationary listener \( = n'' - \nu = \nu \left( {\frac{{c + v}}{{c - v}}} \right) - \nu = \frac{{2\nu v}}{{(c - v)}}\)
Hence option \((a)\, (b)\) and \((c)\) are correct.
${x_1} = a\sin (\omega \,t + {\phi _1})$, ${x_2} = a\sin \,(\omega \,t + {\phi _2})$
यदि परिणामी तरंग में आवृत्ति एवं आयाम अध्यारोपित होने वाली तरंगों के समान हैं तब इनके मध्य कलान्तर होगा
$[\,dB$ માં લબ્ધિ $\left.=10 \log _{10}\left(\frac{ P _{ o }}{ P _{i}}\right)\right]$