Question 323: Wave Optics

Question

$\mathrm{A}$ and $\mathrm{B}$ are two interfering sources where $\mathrm{A}$ is ahead in phase by $54^{\circ}$ relative to B. The observation is taken from point $\mathrm{P}$ such that PB $-$ PA = 2.5 $\lambda$. Then the phase difference between the waves from A and B reaching point P is (in rad)

Accepted Answer

$\begin{aligned}
& 	ext { Total phase difference }=\phi_1+\phi_2 \
& \begin{aligned}
\phi_1 & =54 	imes \frac{\pi}{180}=0.3 \pi \
\phi_2 & =\frac{2 \pi}{\lambda} 	imes(\mathrm{PB}-\mathrm{PA}) \
& =\frac{2 \pi}{\lambda} 	imes 2.5 \lambda=5 \pi \
	herefore \quad \phi_1 & +\phi_2=5 \pi+0.3 \pi=5.3 \pi
\end{aligned}
\end{aligned}$

Answer

$3.5 \pi$

Answer

$5.3 \pi$

Answer

$4.3 \pi$

Answer

$5.8 \pi$

Wave Optics

\mathrm{A}$ and $\mathrm{B}$ are two interfering sources where $\mathrm{A}$ is ahead in phase by $54^{\circ}$ relative to B. The observation is taken from point $\mathrm{P}$ such that PB $-$ PA = 2.5 $\lambda$. Then the phase difference between the waves from A and B reaching point P is (in rad)

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