Hello sir, you mentioned that it must be a monochromatic light source then why did u mention ray of light which means many wavelengths? Also, a single wavelength can not make interference, it must be two (really sorry for the dumb qn. I might be missing something here)

Anytime! Thanks for the comment!

Thanks a lot for the comment! I actually love solving some JEE problems and also some of the entrance exam problems from here. I've put them all in a playlist if it's helpful: kzfaq.info/sun/PLSygKZqfTjPCKus_q7vWTaRClR2hi6XFe

Definitely not a stretch, I have a playlist here of a LOT of PAT problems: kzfaq.info/sun/PLSygKZqfTjPDbpUCdVbr6hjp9bWkO5GvY Also a while ago I did a PAT tips video for Physics Online here: kzfaq.info/get/bejne/bcxgZpylvZrcfp8.html

So for example if you were doing the double slit for sound, the speakers could be swapped. Hope this helps!

@@zhelyo_physics Hi thanks for the reply, wouldn't the pattern be the same if the 2 speakers were the same and swapped though?

will do, thanks!

so if it is a maximum, this can only occur if a peak meets a peak. This can happen with zero path difference, path difference of 1 wavelength, 2 wavelengths etc.. It can't be zero because then we would have the central maximum for the path from both slits to be equal. We consider the next one along which is a path difference of 1 wavelength. Hope this helps! My video on superposition might be helpful too: kzfaq.info/get/bejne/q6moqtaavJ2qY6c.htmlsi=HcOiWuvMT7neli6n

This video is in the works! : ) It's a tricky one to prepare but here are some tips (more to come soon): Paper 3 questions and general physics tips: kzfaq.info/get/bejne/nb2Bo7Wbmqi-Yps.html , kzfaq.info/get/bejne/qNJ3i5mKv7i1j5c.html , kzfaq.info/get/bejne/abGonZupsbaqeIU.html , kzfaq.info/get/bejne/hdCera513r2-g4U.html

hi, it's the equation which only works as the double slit formula uses a small angle approximation and the angle in question is only small when D>>a. Hope this helps!

@@zhelyo_physics this makes sense! Thank you for all your help. Your vdieos have been very helpful for the upcoming exam S

do you know why? or is it a mistake

The phase difference on each individual waves is reversed on each location. Hope this helps! Might do another video on this topic.

anytime! Thanks for the comments!

the phase between the waves is reversed, hope this helps! I might do a further video on this

@@zhelyo_physics Thanks for your response. I am sorry I did not clarify. If the 2 speakers are hooked to the *same* signal source then are they not indistinguishable from one another ? in which case there would be no difference to the interference pattern ?

@@zhelyo_physics Thanks for your response. Sorry Sir, What I meant is that if both (identical) speakers were connected to the same signal (in parallel) in which case they are coherent and indistinguishable from one another. In this case then I would expect no change to the interference pattern right ?

No problem! So we have to consider the path difference, the path of each ray is reversed. Let's say one source had a path of 7m, the other of 8m. If they are reversed, then one will have a path difference of 8, the other of 7. They would still have the same overall magnitude of the path difference but if before a max met a max, now they'd differ by half a wavelength and a max would meet an antimax. I think that's right : ) Hope this helps more!

Yeah ur right, he was suppose to say maxima the next maxima

Funny, I am actually about to record a new video on a very similar topic later today, stay tuned!

Thank you!

Sure! No problem, actually easier to use the rule for 2 angles. So θ1 can be shown to equal θ2. There are two right angles in the triangle just to the right giving us 90-θ1, this is the angle between the gray and blue line on my diagram. Next consider the angle from gray to yellow - this is also 90. So for that angle 90-θ1+θ2=90 i.e. it follows that θ1=θ2, we also have the 90 degree angle so both triangles are similar. Hope this helps!

@@zhelyo_physicswhy does the θ2 calculated using this method equal the angle created between the light wave from slit 2 and the perpendicular line from the slit to the screen? The lines aren’t parallel so alternate angles doesn’t work, surely? The method shown here does show that θ1 = θ2 but I’m unsure about how that connects to the angle we actually need to find.

by path differnce i mean "P"

also how is sin theta 1 approximatley just theta? (sorry fo r these dumb questions)

Excellent questions! So since the point X is a maximum, the path difference my definition is a multiple of the wavelength, i.e. peak meets a peak and the first maximum above the central maximum has a path difference equal to lamda. In terms of angles, in maths the sine of the angle is about equal to the angle itself in radians so we can approximate it. Hope this helps!

@@zhelyo_physics oh ok that makes sense thank you. But also ik this is a stretch but if you could make more videos on some aqa physics topics it would be helpful as you are my main channel for revision. for example in waves topics such as : laser experiment, core cladding( and TIR), intensity graphs for single and double slit are missing.

In the order to produce a maximum, the distance has to be equal to the peak to peak distance which is one wavelength. Hope this helps! : )

@@zhelyo_physics is this because theres constructive interference at a maximum, so the phase difference between the waves has to be one wavelength? or the path difference has to be one wavelength? path or phase ahhh i get confused

path difference, so path difference is measured in m and hence to be equal to the wavelength. Phase difference is measured in degrees or radians with 360 degrees or 2pi radians equal to one oscillation. Hope this helps!

No worries! So if it is not, we can't make the approximations we have done in the proof. In practice the slit separation will virtually always be bigger than the distance to the screen. The slit separation is normally of the order of 0.25 mm or similar on standard equipment. We may still get interference but we just can't apply the double slit equation. Hope this helps!

Is it because you considered the integer “n” to be 1?

Yep, precisely, right!

@@zhelyo_physics understood! Thank you! :)

good luck revising! : )

@@zhelyo_physics I have a question, for AQA, the derivation of double slit is the one where the two lines meet a fringe (the one in this video) What is the difference between that one and the diffraction grating one where the lines are parallel and don't meet at a point?

@@prodcio excellent question! the diffraction grating one applies for a grating i.e. multiple slits (typically of the order of a few hundred per mm)

Hi sir, I am fascinated by all of your videos and you’re really helping me in the process of revising for my mocks. Thank you for your help. I do have just one question, though. If the sources of light / sound are coherent and monochromatic, why would the maxima become minima and vice versa, when they are swapped? 16:11