A) f
B) 2f
C) f/2
D) f/4
Correct Answer: A
Solution :
(i) Here, \[\frac{\lambda }{2}=l\Rightarrow \lambda =2l\] So, \[{{v}_{1}}=\frac{v}{2l}\] (ii) And \[\frac{\lambda }{4}=\frac{\ell }{2}\] \[\lambda =\frac{4l}{2}=2l\] \[\therefore \] \[{{v}_{2}}=\frac{v}{2l},\]the sameYou need to login to perform this action.
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