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In Young's double-slit experiment, monochromatic light of wavelength $\lambda$, the intensity of light at a point on the screen where path difference is $\lambda$ is $K$ units. What is the intensity of light at a point where path difference is $\frac{\lambda}{3} ?$
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Verified Answer
As the resultant intensity at a point,
$$
\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2+2 \sqrt{\mathrm{I}_1 \mathrm{I}_2} \cos \phi
$$
When the path difference $=\lambda$,
phase difference $=0^{\circ}$
$$
\begin{aligned}
\therefore \quad \mathrm{I}_{\mathrm{R}} &=\mathrm{I}+\mathrm{I}+2 \sqrt{\mathrm{I} \times \mathrm{I}} \cos 0^{\circ} \\
&=2 \mathrm{I}+2 \sqrt{\mathrm{I}^2} \times 1=2 \mathrm{I}+2 \mathrm{I}=4 \mathrm{I}=\mathrm{K} .
\end{aligned}
$$
When the path difference $=\frac{\lambda}{3}$, phase difference $\phi=\frac{2 \pi}{3}$
$$
\begin{aligned}
\therefore \quad \mathrm{I}_{\mathrm{R}}^{\prime} &=\mathrm{I}+\mathrm{I}+2 \sqrt{\mathrm{I} I} \cdot \cos \left(\frac{2 \pi}{3}\right) \\
&=2 \mathrm{I}+2 \sqrt{\mathrm{I}^2} \times\left(-\frac{1}{2}\right)=2 \mathrm{I}-\frac{2 \mathrm{I}}{2}=\mathrm{I} \\
\therefore \quad \mathrm{I}^{\prime} &=\frac{\mathrm{K}}{4}
\end{aligned}
$$
$$
\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2+2 \sqrt{\mathrm{I}_1 \mathrm{I}_2} \cos \phi
$$
When the path difference $=\lambda$,
phase difference $=0^{\circ}$
$$
\begin{aligned}
\therefore \quad \mathrm{I}_{\mathrm{R}} &=\mathrm{I}+\mathrm{I}+2 \sqrt{\mathrm{I} \times \mathrm{I}} \cos 0^{\circ} \\
&=2 \mathrm{I}+2 \sqrt{\mathrm{I}^2} \times 1=2 \mathrm{I}+2 \mathrm{I}=4 \mathrm{I}=\mathrm{K} .
\end{aligned}
$$
When the path difference $=\frac{\lambda}{3}$, phase difference $\phi=\frac{2 \pi}{3}$
$$
\begin{aligned}
\therefore \quad \mathrm{I}_{\mathrm{R}}^{\prime} &=\mathrm{I}+\mathrm{I}+2 \sqrt{\mathrm{I} I} \cdot \cos \left(\frac{2 \pi}{3}\right) \\
&=2 \mathrm{I}+2 \sqrt{\mathrm{I}^2} \times\left(-\frac{1}{2}\right)=2 \mathrm{I}-\frac{2 \mathrm{I}}{2}=\mathrm{I} \\
\therefore \quad \mathrm{I}^{\prime} &=\frac{\mathrm{K}}{4}
\end{aligned}
$$
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