Question

In the given figure, $C$ is middle point of lines $S_1{S}_2.$ A monochromatic light of wavelength $\lambda$ is incident on slits. The ratio of intensities of $S_3$ and $S_4$ is

• A. zero
• B. $\infty$
• C. 4 : 1
• D. 1 : 4

Answer 1

Answer: (B)

Path difference, $\Delta x=S_1{S}_3-S_2{S}_3=0$
$\therefore \varphi =\frac{2\pi }{\lambda }\Delta x=0$
$\therefore I_3=I_0+I_0+2\sqrt{I_0+I_0\left(\cos 0^{\circ }\right)}$
$\therefore I_3=4I_0$
The path difference at $S_4$ is
$\Delta x^{\prime }=S_1{S}_4-S_2{S}_4=\frac{xd}{D}$ (Here, $x=\frac{\lambda D}{2d}$)
$=\frac{d}{D}\times \frac{\lambda D}{2d}=\frac{\lambda }{2}$
$\therefore {\varphi }^{\prime }=\frac{2\pi }{\lambda }\cdot \frac{\lambda }{2}=\pi$
$\therefore I_4=I_0+I_0+2I_0\cos \pi =0$
$\therefore$ Ratio $=\frac{I_3}{I_4}=\frac{4I_0}{0}=\infty$