A) 0.7 mm
B) 0.1 mm
C) 0.2 mm
D) 0.5 mm
Correct Answer: D
Solution :
\[\lambda =\frac{v}{n}=\frac{350}{350}=1\,m\]=100 cm Also path difference \[(\Delta x)\] between the waves at the point of observation is \[AP-BP=25cm\]. Hence Þ \[\Delta \varphi =\frac{2\pi }{\lambda }(\Delta x)=\frac{2\pi }{1}\times \left( \frac{25}{100} \right)=\frac{\pi }{2}\] Þ \[A=\sqrt{{{({{a}_{1}})}^{2}}+{{({{a}_{2}})}^{2}}}\]\[=\]\[\sqrt{{{(0.3)}^{2}}+{{(0.4)}^{2}}}\]= 0.5 \[mm\]You need to login to perform this action.
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