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In Young's double slit experiment, the distance between the two slits is $0.1 \mathrm{~mm}$, the distance between the slits and the screen is $1 \mathrm{~m}$ and the wavelength of the light used is $600 \mathrm{~nm}$. The intensity at a point on the screen is $75 \%$ of the maximum intensity. What is the smallest distance of this point from the central fringe?
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The correct answer is:
$1.0 \mathrm{~mm}$
$\mathrm{d}=0.1 \mathrm{~mm}, \mathrm{D}=1 \mathrm{~m}, \lambda=600 \mathrm{~nm}$
$\mathrm{I}_{\mathrm{p}}=75 \%$ of maximum or $\mathrm{I}_{\mathrm{p}}=3 \mathrm{I}_{0}$
Where $I_{0}$ is the intensity of a single wave
now $\mathrm{I}_{\mathrm{P}}=3 \mathrm{I}_{0}=\left(\sqrt{\mathrm{I}_{0}}\right)^{2}+\left(\sqrt{\mathrm{I}_{0}}\right)^{2}+2 \sqrt{\mathrm{I}_{0} \times \mathrm{I}_{0}} \cos \phi$
$\therefore \cos \phi=\cos \frac{\pi}{3}$, also $\Delta \mathrm{x}=\frac{\mathrm{yd}}{\mathrm{D}}$
now $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} \times \frac{\pi}{3}=\frac{\lambda}{6} \quad \therefore \mathrm{y}=\frac{\lambda \mathrm{D}}{6 \mathrm{~d}}=\frac{600 \times 10^{-9} \times 1}{6 \times 0.1 \times 10^{-3}}$ or $\mathrm{y}=1 \mathrm{~m}$
$\mathrm{I}_{\mathrm{p}}=75 \%$ of maximum or $\mathrm{I}_{\mathrm{p}}=3 \mathrm{I}_{0}$
Where $I_{0}$ is the intensity of a single wave
now $\mathrm{I}_{\mathrm{P}}=3 \mathrm{I}_{0}=\left(\sqrt{\mathrm{I}_{0}}\right)^{2}+\left(\sqrt{\mathrm{I}_{0}}\right)^{2}+2 \sqrt{\mathrm{I}_{0} \times \mathrm{I}_{0}} \cos \phi$
$\therefore \cos \phi=\cos \frac{\pi}{3}$, also $\Delta \mathrm{x}=\frac{\mathrm{yd}}{\mathrm{D}}$
now $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} \times \frac{\pi}{3}=\frac{\lambda}{6} \quad \therefore \mathrm{y}=\frac{\lambda \mathrm{D}}{6 \mathrm{~d}}=\frac{600 \times 10^{-9} \times 1}{6 \times 0.1 \times 10^{-3}}$ or $\mathrm{y}=1 \mathrm{~m}$
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