Question

If two waves of equal intensities $I_{1}=I_{2}=I_{0}$, meets at two locations $P$ and $Q$ with path difference $\Delta_{1}$ and $\Delta_{2}$ respectively, then the ratio of resultant intensity at points $P$ and $Q,\left(\frac{I_{P}}{I_{Q}}\right)$ will be

• A. $\frac{\cos ^{2}\left(\frac{\Delta_{1}}{\lambda}\right)}{\cos ^{2}\left(\frac{\Delta_{2}}{\lambda}\right)}$
• B. $\frac{\cos ^{2} \Delta_{1}}{\cos ^{2} \Delta_{2}}$
• C. $\frac{\cos ^{2}\left(\frac{\pi \Delta_{1}}{\lambda}\right)}{\cos ^{2}\left(\frac{\pi \Delta_{2}}{\lambda}\right)}$
• D. $\frac{\Delta_{1}}{\Delta_{2}}$

Answer 1

Answer: (C)

Resultant intensity is given by $I=4 I_{0} \cos ^{2} \phi / 2$
Now, phase difference $(\phi)=\frac{2 \pi}{\lambda} \times \Delta$ (path difference)
As a path difference of one wavelength corresponds to a phase difference of $2 \pi$ radius.
$\Rightarrow I=4 I_{0} \cos ^{2}\left(\frac{2 \pi \Delta}{2 \lambda}\right)$
$=4 I_{0} \cos ^{2}\left(\frac{\pi \Delta}{\lambda}\right)$
$\therefore \frac{I_{P}}{I_{Q}}=\cos ^{2}\left(\frac{\pi \Delta_{1}}{\lambda}\right) / \cos ^{2}\left(\frac{\pi \Delta_{2}}{\lambda}\right)$