Chapter 46: Problem 6

Question: A plane wave with wavelength $\lambda~=~593~{\rm nm}$ falls on a slit of width $a~=~420~{\rm\mu m}$. A thin converging lens having a focal length of $f~=~71.4~{\rm cm}$ is placed behind the slit and focuses the light on a screen. Find the distance on the screen from the center of the pattern to the second minimum.

Solution: The angle at which the second diffraction minima occurs is determined by $a\sin\theta_2~=~2 \lambda$, from which we find that $\theta_2~=~0.16^o$. The angle of the intensity pattern is unmodified by the lens, except that parallel light is focused a distance f from the lens. This the distance between the second minimum and the center of the central maxima is $h~\sim~\theta f$, for small $\theta$, which gives $h~=~(71.4)({0.16\pi\over 180})~=~2~{\rm mm}$.