Is it possible to derive this result from the expression for the displacement produced by a single slit? I mean A is proportional to sin(½ka*sin(theta))/½ka*sin(theta) where a is a width of the slit I tried but got something really different. However, double slit is clearly just a superposition of two slits Btw great video!
It is indeed, but note that the result you quoted applies to a slit of finite width while I'm assuming in this video that the slits are infinitesimally thin. The full analysis is mostly the same as in the video, except |A| is no longer a constant and should really be replaced by your expression involving the sinc function. So the phasor diagram I drew is still valid, but you have to imagine that the length of the two phasors we're adding changes as θ varies. That means the full solution is really the product of the expression in the video and the sinc function. In the limit a -> 0, the single-slit part of the expression tends to a constant independent of θ, because sinc(0) = 1, and you recover the result from the video.
More optics content please!
Is it possible to derive this result from the expression for the displacement produced by a single slit? I mean A is proportional to sin(½ka*sin(theta))/½ka*sin(theta) where a is a width of the slit
I tried but got something really different. However, double slit is clearly just a superposition of two slits
Btw great video!
It is indeed, but note that the result you quoted applies to a slit of finite width while I'm assuming in this video that the slits are infinitesimally thin. The full analysis is mostly the same as in the video, except |A| is no longer a constant and should really be replaced by your expression involving the sinc function. So the phasor diagram I drew is still valid, but you have to imagine that the length of the two phasors we're adding changes as θ varies. That means the full solution is really the product of the expression in the video and the sinc function. In the limit a -> 0, the single-slit part of the expression tends to a constant independent of θ, because sinc(0) = 1, and you recover the result from the video.
@@DrBenYelverton makes sense, thank you!