Single Slit Diffraction - Physics Problems

This physics video tutorial provides a basic introduction into single slit diffraction. It explains how to calculate the width of the central bright fringe and the angular width in degrees given the separation distance of the single slit, the wavelength of light used, and the distance between the slit and the screen.

Ray Diagrams:

The Huygens Principle:

Young's Double Slit Experiment:

Single Slit Diffraction:

Diffraction Grating Problems:

Polarization of Light Problems:

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Brewster's Angle:

Planck's Constant and Blackbody Radiation:

Photon Momentum and Effective Mass:

Wien's Law:

Compton Effect & Wavelength:

Physics 2 Final Exam Review:

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Final Exams and Video Playlists:

Full-Length Videos and Worksheets:

[DudeWhoSaysDeez]

You have a video for every topic that I am ever stuck on. Thank you 🙏🏻

[nknm]

Thank you so much! These videos are *literally* a lifesaver. Every time I turn to this channel it has exactly what I'm looking for.

[minimarsed]

english isnt even my native language but this is what im resorting to because i have no clue what my teacher is saying, thank you for this man

[katlegokokoana]

God bless you bro... you just made prepared for the physics test.

[__ICT__]

U taught this topic better than my actual teacher, thanks

[bennettgrado]

Everyone asking about the dark fringes being a half, it’s because when you calculate the slits, for single slits we divide the width of the split by 2, so for the dark fringes it’s d/2 sin(theta) = 1/2 +)lambda, you multiply the 2 over making d sin(theta) =2 lambda for the first DARK fringe, it’s confusing because the double slits is d sin(theta) =mlamda for the contructive

[pikachu]

At 9:31 The angle (Theta1) should be in degrees and not meters.

[eshi]

you should make a video on thin film diffraction !

[ManzilSafarAurBaateinOfficial]

Please answer 🙏
Consider a slit of width a producing a diffraction pattern, if 'm' is either positive or negative integer than diffraction minima occur under the conditions
1- M-lamda = a sin
2- 2M lambda = a sin
3- m lambda = 2 a sin
4- (2m+1) lambda = 2 a sin

[mashanatanzon]

On question one, shouldn;t the answer be 0.952m? We know that theta = lambda/slit width. And that lambda is = opposite over adjacent, making Resulting in width = 0.952m. Thank you for clarifying this!

[alainfabrice]

I almost left without liking the video 😅👍

[amnaumar]

for dark fringe, shouldn't m be a decimal value like 0.5, 1.5, 2.5 etc?

[xXishibashiX]

This video gives me but a "whiff" of hope

[TheLover]

I thought since the angle is bigger than 10 degrees then we cannot use the formula
y1= Ltan(theta) because it may be used for small angles.
Therefore, wouldn’t we use the formula y=L(m+0.5)lambda/d to find y1? If so, then the distance of the central maximum would be equal to 1.43m.

[BlazinblaiserVods]

Why do we do this is Radians and not Degrees when finding the angle?

[schenzur]

9:33 why aren't we multiplying theta by 2

[MegaTiffanym]

Why didn't you convert the 1.6 cm to meters to be consistent with the rest of the units?

[ethanhope]

why does destructive interference happen on single slit?

[blackart]

Sir how to relate wavelength and intensity in single slit

[nemuirostorageroom]

spent 4 hours watching my prof's lecture, and I feel like I have a brain fog. Literally 1 minute into the video (the part where he says the amplitude of the bright fringe decreases progressively), I went