A) are half as numerous;
B) are one quarter as numerous
C) each carry one quarter of their previous energy
D) each carry one quarter of their previous momentum.
Correct Answer: B
Solution :
\[\Upsilon =A{{\cos }^{2}}\left( 2\pi \,n\,t-2\pi \frac{x}{\lambda } \right)\] \[{{\cos }^{2}}\theta =2d{{\cos }^{2}}\theta -1\] \[{{\cos }^{2}}\theta =\left( \frac{\cos 2\theta +1}{2} \right)\] \[\therefore \] \[\Upsilon =A\left[ \frac{\cos 2\left( 2\pi \,nt-2\pi \frac{x}{t} \right)+1}{2} \right]\] \[\Upsilon =\frac{A}{2}\cos \left[ 2\pi (2n)t-2\pi \left( \frac{2x}{\lambda } \right) \right]+1/2\] amplitude\[=A/2,\] frequency \[=2n\] wavelength \[=\lambda /2\]
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