Question 204: Waves Optics

Question

In a double slit experiment shown in figure, when light of wavelength $$400 \mathrm{~nm}$$ is used, dark fringe is observed at $$P$$. If $$\mathrm{D}=0.2 \mathrm{~m}$$, the minimum distance between the slits $$S_1$$ and $$S_2$$ is _________ $$\mathrm{mm}$$.

Accepted Answer

Path difference for minima at $$\mathrm{P}$$
$$\begin{aligned}
& 2 \sqrt{\mathrm{D}^2+\mathrm{d}^2}-2 \mathrm{D}=\frac{\lambda}{2} \
& 	herefore \sqrt{\mathrm{D}^2+\mathrm{d}^2}-\mathrm{D}=\frac{\lambda}{4} \
& 	herefore \sqrt{\mathrm{D}^2+\mathrm{d}^2}=\frac{\lambda}{4}+\mathrm{D} \
& \Rightarrow \mathrm{D}^2+\mathrm{d}^2=\mathrm{D}^2+\frac{\lambda^2}{16}+\frac{\mathrm{D} \lambda}{2} \
& \Rightarrow \mathrm{d}^2=\frac{\mathrm{D} \lambda}{2}+\frac{\lambda^2}{16} \
& \Rightarrow \mathrm{d}^2=\frac{0.2 	imes 400 	imes 10^{-9}}{2}+\frac{4 	imes 10^{-14}}{4} \
& \Rightarrow \mathrm{d}^2 \approx 400 	imes 10^{-10} \
& 	herefore \mathrm{d}=20 	imes 10^{-5} \
& \Rightarrow \mathrm{d}=0.20 \mathrm{~mm}
\end{aligned}$$

Waves Optics

In a double slit experiment shown in figure, when light of wavelength $400 \mathrm{~nm} is used, dark fringe is observed at P. If \mathrm{D}=0.2 \mathrm{~m}, the minimum distance between the slits S_1 and S_2 is _________ \mathrm{mm}$.

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