When two light waves each of amplitude 'A' and having a phase difference of $\frac{\pi}{2}$ superimposed…
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
$\frac{\mathrm{A}}{\sqrt{2}}$
$2A$
$\sqrt{2} \mathrm{~A}$
$\frac {A}{2}$
Solution
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 $\phi=90^{\circ}$
$\therefore \quad \mathrm{R}=\sqrt{\mathrm{A}^2+\mathrm{A}^2+2 \mathrm{~A}^2 \cos 90^{\circ}}=\sqrt{2 \mathrm{~A}^2}=\sqrt{2} \mathrm{~A}$
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