Problem

Problem

The maximum intensity in Young's double-slit experiment is I0I_0 . Distance between the slits is d=5λd = 5\lambda , where λ\lambda is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance D=10dD = 10d ? (1) I02\frac{I_0}{2} (2) 34I0\frac{3}{4}I_0 (3) I0I_0 (4) I04\frac{I_0}{4}
Intuitive explanation

The key is to find the path difference from each slit to the observation point directly in front of one slit, then use it to determine the phase difference and resulting intensity. When you're on the axis of one slit, that slit contributes a strong signal, but the other slit is offset by distance dd, creating a specific path difference that reduces the total intensity through interference. Calculate this path difference using geometry, convert it to a phase difference, and apply the two-slit intensity formula I=I0cos⁡2(Δϕ/2)I = I_0 \cos^2(\Delta\phi/2) to find how much the waves reinforce or cancel.

Intuitive explanation

The key is to find the path difference from each slit to the observation point directly in front of one slit, then use it to determine the phase difference and resulting intensity. When you're on the axis of one slit, that slit contributes a strong signal, but the other slit is offset by distance dd, creating a specific path difference that reduces the total intensity through interference. Calculate this path difference using geometry, convert it to a phase difference, and apply the two-slit intensity formula I=I0cos⁡2(Δϕ/2)I = I_0 \cos^2(\Delta\phi/2) to find how much the waves reinforce or cancel.

Browse the problem library