An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is $1.70 \mathrm {~m}$ from the slits.If the slits are illuminated with coherent light of wavelength 633 nm,how far apart should the slits be?

Question

Quizplus · Accepted Answer

The distance between bright fringes in a double-slit experiment is given by the formula $y = \frac{\lambda L}{d}$, where $y$ is the fringe separation, $\lambda$ is the wavelength of the light, $L$ is the distance from the slits to the screen, and $d$ is the distance between the slits. Rearranging this formula to solve for $d$, we get $d = \frac{\lambda L}{y}$. Substituting the given values ($\lambda = 633 \times 10^{-9} \, \mathrm{m}$, $L = 1.70 \, \mathrm{m}$, and $y = 15.7 \times 10^{-3} \, \mathrm{m}$), we find $d = \frac{633 \times 10^{-9} \times 1.70}{15.7 \times 10^{-3}} \approx 68.5 \times 10^{-6} \, \mathrm{m}$ or $68.5 \, \mu\mathrm{m}$.

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An Optical Engineer Needs to Ensure That the Bright Fringes $1.70 \mathrm {~m}$