Question

A monochromatic light source of wavelength $\lambda$ is placed at $S$. Three slits $S_1,S_2$ and $S_3$ are equidistant from the source $S$ and the point $P$ on the screen $S_{1}P-S_{2}P=\frac{\lambda }{6}and{S}_{1}P-S_{3}P=\frac{2\lambda }{3}$.
If $I$ be the intensity at $P$ when only one slit is open, the intensity at $P$ when all the three slits are open is

• A. $3I$
• B. $5I$
• C. $8I$
• D. Zero

Answer 1

The correct option is A $3I$

At $P$ the intensity is due to the superpositon of the three waves. Phase difference between $1$ and $2$ is
${\varphi }_{12}=\frac{2\pi }{\lambda }\times 2\times \frac{\lambda }{6}=\frac{2\pi }{3}={120}^{\circ }$

Phase difference between $1$ and $3$ is
${\varphi }_{13}=\frac{2\pi }{\lambda }\times 2\times \frac{2\lambda }{3}=\frac{8\pi }{3}=2\pi +\frac{2\pi }{3}={120}^{\circ }$

$\Rightarrow \left(2\right)\text{and}\left(3\right)$ are in phase
Since superposition is similar to adding vectors,


If $A$ is amplitude of each wave,
Resultant Intensity is
$I_{res}\propto A^2+(2A{)}^2+2A(2A\left)\cos \right({120}^{\circ })I_{res}\propto 3A^2\Rightarrow I_{res}=3I.$