Question

In the given figure, $P$ and $Q$ are two equally intense coherent sources, which are emitting radiations of wavelength $20m$. The separation between $P$ and $Q$ is $5.0$ meter and phase of $P$ is ahead of the phase of $Q$ by ${90}^0$. $A,B,C$ are three distant points of observation, equidistant from the midpoint of $PQ$. The intensity of radiation at $A,B$ and $C$ will be in the ratio of:

• A. $0:1:4$
• B. $4:1:0$
• C. $0:1:2$
• D. $2:1:0$

Answer 1

The correct option is C $0:1:2$

Intensity at a point in double slit interference is $I=4I_0{\cos }^2\frac{\delta }{2}$ where $I_0$ is the intensity from each source and $\delta$ is the phase difference between the waves at the observation point.

Point A:
path difference $\Delta =PQ=5$
Wavelength $\lambda =20$
phase difference $\delta =\frac{2\pi }{\lambda }\Delta +\frac{\pi }{2}=\frac{2\pi }{20}\times 5+\frac{\pi }{2}=\pi$
Hence, point A will have a minima, since condition for minima is $\delta =(2n+1)\pi$

Point C:
The phase difference from P and Q are opposite since P is ahead of Q by $\frac{\pi }{2}$ and since path length from Q to C has an extra $\frac{\lambda }{4}$, it adds a phase of $\frac{\pi }{2}$ both will cancel
Point C will have a maxima and intensity will be $4I_0$ because $\delta =2n\pi$

Point B:
$\delta =\frac{\pi }{2}$

$I=4I_0{\cos }^2\frac{\pi }{2\times 2}=2I_0$

Hence $I_A:I_B:I_C=0:2I_0:4I_0=0:1:2$