Maxima in Single slit diffraction | Wave optics | Physics | Khan Academy

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Let's figure out the angular position of maximas in Single slit diffraction!
00:16 Condition for destructive interference
03:04 Dividing the slit into three parts
06:30 Condition for constructive interference
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Created by Vibhor Pandey

Опубликовано:

12 ноя 2022

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Комментарии : 34

@vikrantkirar 8 месяцев назад

this level of clarity is nowhere found on youtube. Thank you for solving the doubts we didn't even have but needed.

@basicallyapotato5410 7 месяцев назад

omg vikrant sirr your content is great

@drakedakes 2 месяца назад

But it is also slightly untrue. This means that if you go to infinity you cannot tell whether there is odd or even number of wavelet sources. The real reason for diffraction pattern being based on these equations is because average distance of all sources to a point on the screen is equal to the distance of a single point in the very center of the slit to the point on the screen. If you have an access to Crawford's "Waves" there is full explanation.

@joshnirohina318 Год назад

im crying,,,this is beautiful

@Great_Himalayan_ASMR Год назад

why!

@joshnirohina318 Год назад

@@Great_Himalayan_ASMR cuz I remember not understanding diffraction no matter how much I read it and I loved wave optics.... After I found the meaning of diffraction, I can't help but see beauty in it

@Great_Himalayan_ASMR Год назад

@@joshnirohina318 share ur contact with me

@rizzmechanics Год назад

@@Great_Himalayan_ASMRbro what😂

@xenongod1823 9 месяцев назад

how

@unnicheethedath1988 Год назад

Khan academy always saves lives❤❤

@techexplained8289 7 месяцев назад

Thank you so much sir

@sparespare1146 Год назад

Saved my life fr

@xenongod1823 9 месяцев назад

@yujintan7286 4 месяца назад

This video does clarify my doubts on how can constructive interference occur when we divided the slit in odd number sections thanks for your work sir, but does all the tiny section left on the slit contribute to constructive interference or just some portion of it?(eg , (1/3)a ) , if it does, how can we prove it?

@rohakgupta7066 Год назад

you have shown a maxima at 3lambda/2 but shouldn't destructive interference lead to a minima instead of maxima

@joshnirohina318 Год назад

thats cuz there's still one part that does not destructively interfere and hence does not get cancelled and so contributes to the maxima

@shubham_bandhavakar8826 6 месяцев назад

but why that one part@@joshnirohina318

@tuongdihoc Год назад

How about if n = 0, so a.sin(theta) = lamda/2 --> is this the maxima at the center?

@MohammadAbdullah-dg1rp 8 месяцев назад

Yes but n is a natural no. and starts from 1 so it cannot be 0 :)

@monowaraalibithi3421 Год назад

Isn't "n lembda" the condition for constructive interference?

@joshnirohina318 Год назад

thats for ydse...this is diffraction...u may ask why and thats exactly wat this vid explains

@imostliked2277 Год назад

i love you

@iit864 Год назад

Too

@xenongod1823 9 месяцев назад

@LASTCHANCE-mi8zy 4 месяца назад

💋💋💋💋💋

@abhinavkashyap9565 2 месяца назад

Love u too bro

@alexdvornikov2056 2 месяца назад

The formula derived in this video is not exact. It is an approximation. There is no simple expression for single-slit maxima. One must numerically solve a transcendental equation for the intensity profile to find the maxima. The numerical solution is close but not exactly what is shown in this video.

@nomann5244 Год назад

0:23 from where we get the condition for 1st minima. should we memorize it? is this the education system where we learn to memorize without understanding? This is unexpected from khan acedemy👎