Question 327: Waves 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

$$5.8 \pi$$

Waves 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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