Question 10: Wave Optics

Question

In the given figure, $$C$$ is middle point of line $$S_1 S_2$$. A monochromatic light of wavelength $$\lambda$$ is incident on slits. The ratio of intensities of $$S_3$$ and $$S_4$$ is

Accepted Answer

$$\begin{aligned}
& 	ext { At } S_3, \Delta x=S_1 S_3-S_2 S_3=0 \
& 	herefore \quad \phi=\frac{2 \pi}{\lambda} \square \Delta x=0 \
& \quad I_3=I_0+I_0+2 \sqrt{I_0 	imes I_0} \cos 0^{\circ} \
& \Rightarrow \quad I_3=4 I_0
\end{aligned}$$
The path difference at $$S_4$$ is
$$\begin{aligned}
\Delta x^{\prime} & =S_1 S_4-S_2 S_4 \
& =\frac{d y}{D}=\frac{d}{D} 	imes \frac{\lambda D}{2 d}=\frac{\lambda}{2} \quad\left(\because y=\frac{\lambda D}{2 d}ight) \
	herefore \quad \phi^{\prime} & =\frac{2 \pi}{\lambda} 	imes \frac{\lambda}{2}=\pi \
\Rightarrow I_4 & =I_0+I_0+2 I_0 \cos \pi=0 \Rightarrow \frac{I_3}{I_4}=\frac{4 I_0}{0}=\infty
\end{aligned}$$

Answer

0

Answer

$$\infty$$

Answer

4 : 1

Answer

1 : 4

Wave Optics

In the given figure, $C is middle point of line S_1 S_2. A monochromatic light of wavelength \lambda is incident on slits. The ratio of intensities of S_3 and S_4$ is

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