no worries, honestly, any help needed and i will be happy to help . erm firm choice (top choice) King's College London (AAA) and insurance (back up) Queen Mary (BBB).
lets say the distance between the centre of 2 adjacent slits is 'd' which can be worked out by doing no.of.slits.per.metre1. so as the wavefront of the source comes in, it diffracts at an angle θ towards the screen, this can be calculated by using a protractor or a spectrometer. if you get constructive interference you know that the path difference for say the first order is λ. now using some trig, we know that sinθ=hypotenuseopposite. sub in λ and 'd' to give sinθ=dλ. which gives dsinθ=λ. now for every maxima or order on the screen, that means you are getting constructive interfence, therefore the path difference with always have to be a whole number (integer) multiple of λ, so lets call that nλ. therefore you get dsinθ=nλ
on the diagram A to C represent λ
Sweet, thank you for that explanation, in my book, I can follow it up to the dsin(theta) part, but I don't understand how to explain the part with the n in words, and you've just done that, thanks a lot buddy!
Sorry, I just ate my crappy chicken wraps. I am very confident for the exam. Just had ICT exam today. And, sorry, I have never posted anything on the biology section chemistry section. But yeah, maths physics a lot of times.