Wave Optics Part 2

Concept:In Young's double slit experiment, the intensity at any point is given by the interference formula for two coherent waves.Explanation:Intensity from each slit is $I_0$.Given: $d = 2\ \text{mm} = 2 \times 10^{-3}\ \text{m}$, $D = 10\ \text{m}$, $\lambda = 6000\ \text{Å} = 6 \times 10^{-7}\ \text{m}$.The point directly in front of one slit is at a distance $y = d/2$ from the central maximum.Path difference: $\Delta x = \frac{d y}{D} = \frac{d(d/2)}{D} = \frac{d^2}{2D}$.Substitute: $\Delta x = \frac{(2 \times 10^{-3})^2}{2 \times 10} = \frac{4 \times 10^{-6}}{20} = 2 \times 10^{-7}\ \text{m}$.Phase difference: $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{6 \times 10^{-7}} \times (2 \times 10^{-7}) = \frac{4\pi}{6} = \frac{2\pi}{3}$.Resultant intensity: $I = 4I_0 \cos^2(\phi/2) = 4I_0 \cos^2(\pi/3)$.$\cos(\pi/3) = 1/2$, so $I = 4I_0 \times (1/2)^2 = 4I_0 \times 1/4 = I_0$.Thus the intensity equals that of a single slit.Answer:Option B: $I_0$