A) \[\frac{c}{c-v}\nu \]
B) \[\frac{c}{c-2v}\nu \]
C) \[\frac{2v}{c}\nu \]
D) \[\frac{2c}{v}\nu \]
Correct Answer: B
Solution :
In this case, we can assume as if both the source and the observer are moving towards each other with speed v. Hence \[\nu '=\frac{c-{{u}_{o}}}{c-{{u}_{s}}}\nu =\frac{c-(-v)}{c-v}\nu =\frac{c+v}{c-v}\nu \] \[=\frac{(c+v)(c-v)}{{{(c-v)}^{2}}}\nu =\frac{{{c}^{2}}-{{v}^{2}}}{{{c}^{2}}+{{v}^{2}}-2vc}\nu \] Since v<<c, therefore \[\nu '=\frac{{{c}^{2}}}{{{c}^{2}}-2vc}=\frac{c}{c-2v}\nu \]
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