Question

Path difference @ganeshie8

Answer 1

@ganeshie8 @hartnn @vishweshshrimali5

Answer 2

Was that an attempt at making a harp?

Answer 3

@astrophysics Yeah physics give you cravings for weird stuff

Answer 4

Why is the path difference not zero when these wavefronts pass through the diffraction grating?

Answer 5

I don't understand your question, what's going on in this scenario? What paths are we looking at and what difference are we looking for?

Answer 6

It's a conceptual waves question not a calculatory one

Answer 7

Okay so this is the question. The answer is formula assumes no path difference of light before entering the grating but there is some path difference before the light enters the grating.
D = separation of slits nLembda =path difference

Answer 8

So how's the path difference not zero?

Answer 9

Oh interesting I don't know. Perhaps the wavelength is much shorter than the diffraction spacing, so it's not diffracting, only reflecting?

Answer 10

No otherwise it would've been mentioned in the answer. The answer only states for path difference

Answer 11

Oh ok I see I misinterpreted the drawing.

I'm pretty sure the reason you can't use \(d \sin \theta = n \lambda\) is because the incident light isn't perpendicular to the diffraction grating.

Answer 12

Ahh ok that's what it says now that I have reread the start of it, so now you want to know why that'd mess it up eh?

Answer 13

Yes but still they say there is a path difference. WHere is that path difference?

Answer 14

oh got it this is ome consequence of geomtery

Answer 15

Although in terms of this formula, it might not quite make sense. We could probably fix this though, but the point is the two pairs of light are in phase I'm pretty sure