Question

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{A}{\sqrt{2
• B. \sqrt{2}A
• C. 2A
• D. zero

Hint:

When two vectors of equal magnitude $A$ are at $90^{\circ}$, their resultant is always $\sqrt{2}A$.

Answer 1

The Correct Option is A

Concept:
When two harmonic waves of amplitudes $A_1$ and $A_2$ superpose with phase difference $\phi$, the resultant amplitude is: \[ A_r=\sqrt{A_1^2+A_2^2+2A_1A_2\cos\phi} \] This is obtained using vector addition (phasor method).
Step 1: Given values
Both waves have equal amplitudes: \[ A_1=A,\qquad A_2=A \] Phase difference: \[ \phi=\frac{\pi}{2} \]
Step 2: Substitute into formula
\[ A_r=\sqrt{A^2+A^2+2(A)(A)\cos\frac{\pi}{2 \]
Step 3: Use value of cosine
\[ \cos\frac{\pi}{2}=0 \] So, \[ A_r=\sqrt{A^2+A^2} \] \[ A_r=\sqrt{2A^2} \]
Step 4: Simplify
\[ A_r=\sqrt{2}\,A \]
Step 5: Final Answer
The resultant amplitude is: \[ \boxed{\sqrt{2}\,A} \] Quick Tip:
For equal amplitudes:

• $\phi=0 \Rightarrow 2A$
• $\phi=\pi \Rightarrow 0$
• $\phi=\frac{\pi}{2} \Rightarrow \sqrt{2}A$