Hi, at 21:07, you are using the formula \theta= 1.22lambda/b, but I don't understand why \theta is s/d. And thank you for these amazing videos it genuinely helps a lot.
theta is s/d as the triangle is a right angle triangle, take the opposite over the adjacent and that is sin(theta), for small theta it is equal to sin(theta)
Hi I would like to ask, theoretically the resolution angle is constant, or does it change if the two source points are not at infinity but at a short distance from the lens?
Hello Mr. Doner, It came to my attention that a portion of your older "free" videos (before the membership feature) were listed as Members-only. I wanted to ask why did this change occur, as I usually revise for exams using your physics videos since a little more than a year. I thought only the new videos you made would be for members, not the old free ones from 2013.
Some of the new videos were set to public (i.e. Doppler Effect and Resolution) I made individual decisions for other videos (i.e. some videos could not be monetized, so I switched them to member's only videos so that they could help to garner some income.) Most of the old videos remain public but there are likely to be some inconveniences. I am now retired and have a wife and disabled daughter to care for, so this income has become important to me.
It is a small angle approximation that the triangle is essentially a sector of circle, and the definition of an angle in radians is the arc length divided by the radius.
it is easier to make... angular separation of our maxima bigger than something that's small than it is to make it bigger than something that's a little bit bigger
you are SOOO saving me rn
Thank you so much for your videos. Such a help with this course
Glad it was helpful!
Youre the best
Hi, at 21:07, you are using the formula \theta= 1.22lambda/b, but I don't understand why \theta is s/d. And thank you for these amazing videos it genuinely helps a lot.
theta is s/d as the triangle is a right angle triangle, take the opposite over the adjacent and that is sin(theta), for small theta it is equal to sin(theta)
@@yahyahammoudeh1944 Thanks! so it's the small angle approximation.
Hi I would like to ask, theoretically the resolution angle is constant, or does it change if the two source points are not at infinity but at a short distance from the lens?
Great video! Thanks!
Glad you liked it!
Hello Mr. Doner,
It came to my attention that a portion of your older "free" videos (before the membership feature) were listed as Members-only.
I wanted to ask why did this change occur, as I usually revise for exams using your physics videos since a little more than a year. I thought only the new videos you made would be for members, not the old free ones from 2013.
Some of the new videos were set to public (i.e. Doppler Effect and Resolution) I made individual decisions for other videos (i.e. some videos could not be monetized, so I switched them to member's only videos so that they could help to garner some income.) Most of the old videos remain public but there are likely to be some inconveniences. I am now retired and have a wife and disabled daughter to care for, so this income has become important to me.
I'm sorry to hear that Mr. Doner. I hope you get the recognition you deserve. Keep up the great work 💪
Thanks!
How come at 29:31 we are using the second order maximum? how do we know that? I get that 28.3 is greater than 2, but why 2?
We need for mN to be larger than R=998. N=900, so the lowest order maximum that will resolve is m=2.
At around 20:09, why does s/d equal theta? It's obviously not a right triangle.
It is a small angle approximation that the triangle is essentially a sector of circle, and the definition of an angle in radians is the arc length divided by the radius.
@@donerphysics thank you!
I love you
it is easier to make... angular separation of our maxima bigger than something that's small than it is to make it bigger than something that's a little bit bigger
I love u