MHT CET · Physics · Oscillations
When two light waves each of amplitude 'A' and having a phase difference of \(\frac{\pi}{2}\) superimposed then the amplitude of resultant wave is
- A \(\frac{\mathrm{A}}{\sqrt{2}}\)
- B \(2A\)
- C \(\sqrt{2} \mathrm{~A}\)
- D \(\frac {A}{2}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{2} \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
The formula for resultant amplitude is,
\(R=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \phi}\)
Here, \(A_1=A_2=A\) and \(\phi=90^{\circ}\)
\(\therefore \mathrm{R}=\sqrt{\mathrm{A}^2+\mathrm{A}^2+2 \mathrm{~A}^2 \cos 90^{\circ}}=\sqrt{2 \mathrm{~A}^2}\) \(=\sqrt{2} \mathrm{~A}\)
\(R=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \phi}\)
Here, \(A_1=A_2=A\) and \(\phi=90^{\circ}\)
\(\therefore \mathrm{R}=\sqrt{\mathrm{A}^2+\mathrm{A}^2+2 \mathrm{~A}^2 \cos 90^{\circ}}=\sqrt{2 \mathrm{~A}^2}\) \(=\sqrt{2} \mathrm{~A}\)
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