Physics - Optics: Lenses (1 of 4) Converging Lens
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- Опубліковано 14 жов 2024
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In this video I will show you how to find the location of the image when the object is placed 100cm away from the converging lens.
Why can't you be my physics teacher.. You explained this perfectly!
We did these at school and I got them down, then we moved onto magnifying equations for a week (the 1/f= 1/do + 1/di ) and completely forgot how to do these
Thank god for this guy!!!!!!
Brilliant! I was confused at school until i watched this video :)
You got me through physics 1 and 2. Might even go round 3
Keep it going
God bless these teachers
One of the Great teacher I subscribe to your Channel to Enhanced my Physics Understanding! Absolutely I've Learned So much Thank you..
Do you have details about camara obscura concept? Thanks , great videos.
Thank you so much for your great help , I just have a question that confuse me , in vergence formula , U+D=V
where U is the vergence of object ray entering lens , D vergence added by the length (lens power in diopter) and V vergence of image ray living lens
If I understand this right , it seems that
D=1/f , V=1/s' , U = 1/s
So 1/s + 1/f = 1/s'
Than 1/f = 1/s' - 1/s , how is that possible , are this 2 formula unrelated and I 'm just confused , thank you in advance .
thank you for the lessons professors. are you planning to make videos about knowledge that olympiad students of high school should have? that'd be really helpful.
u need to giveup
Your videos are awesome! Yet I can't help but wonder if my clarity comes from the fact that I am able to sort ideas that I was introduced in lecture... Plus all viewers have the luxury of playing certain segments back over again, which differences from ciphering ones lecture notes... I guess my conclusions is that learning requires comprehension skills and out of class effort that everyone here is putting in by watching your videos, so thank you very much Michel.
I notice that in one of the previous videos you said a convex mirror has its focal length negative, why is it that the converging lens which has a convex shape is having a positive focal length in this video?
The answer for that can be found in this playlist on the lens equation: PHYSICS 55.2 THE LENS EQUATION
now thats what i call a real teacher , we took this yesterday in class with mirror and types of it and lens in general he didnt explain to us that their is thick and thin lens and i can tell you he made a bad job because i didnt understand a word from him, other than he is using different langauge 'other then english' and about the rule he gave us 1/f = 1/p +1/p' and i didnt knew you mulitiply the two numbers on top and where you said the point intersection he said A'B' ur awesome can you give us more? im going threw alot and i have officail (goverment) Exams starts in 19/6/2017
They are in these playlists: PHYSICS 55 CONVERGING AND DIVERGING LENSES PHYSICS 55.1 LENSES AND MIRRORS UNDERSTOOD
Michel van Biezen sir yes sir i will see them :)
Wow how does he draw straight lines free hand ????😯
swear to god I thought the same thing
can you explain why you drew the lines the way you did? how do you know for sure that the light hits the focus? don't we consider the trend?
Sorry for being late, but for all intents and purposes it always hits F.
তুই একটা বস।।
You are really a boss
nice videos now i got a clarity about lenses and mirrors
Your video is understanding can you explain galileon transformation
Oh Michel this is really good! At least now I know that negative means the image is either inverted or virtual.
I finally understand lenses. Thank you!
This guy is a CHAMP
In all the optic problems, how can we tell whether the focus is negative or positive?
Thanks for the great explanation, Professor Beizen.
Do you have any plans to make lectures about Fourier Optics?
That is part of the plan, but we first have to finish Fourier Series, Fourier Transforms, and thick lenses
Excellent video. Thank you!
When it is at 15cm, Sprime at -37.5cm, magnified by 2.5times.
What's Hamilton canonical equation of this oscillator?
A bit our of context.
Your mental math skills are too good :P, any tips on how you do calculations like 2500/75 in your head!?
Abdullah,
I grew up before calculators existed.
2500 / 75 both numerator and denominator are divisible by 25
= 100/3 and that can be easily simplified to 33.33
Michel van Biezen Ah, gocha! I gotta discover some more of these little mental math tricks, i'm way too calculator dependent. Anyways, thanks again for the very helpful lectures!
***** I've actually been practicing since this comment...soooooo...yeaaaah...looks like you're alone on this one Nataly :P xD :D
thanks from iraq 🇮🇶💙
Glad you liked it. Welcome to the channel! 🙂
@@MichelvanBiezen Can I have your account? because I have a lot of questions. If you have time ☺️
You can pose your questions here, which is easier for me to respond to (when I have the time) rather than my e-mail account
on.. f.s1/s1- f can the s1 be replaced by the height of the object?
Do you have a video where you derive the formulas you have on the side there? Great video as always; if not, I can try Googling it.
Gregory Serrano Are you asking how to go from 1/f = 1/S+ 1/S' to S' = Sf / (S-f) ?
First re-arrange the equation
1/S' = 1/f - 1/S
Then find the common denominator
1/S' = S/Sf - f/Sf
then combine
1/S' = (S - f) / Sf
then invert
S' = Sf / (S - f)
Outstanding! Thank you so much! I just went through your playists. They consist of almost every kind of science or math I am familiar with. I wonder how many terabytes of information are in your head? :-) Thanks for all your help.
Michel van Biezen Hello! I was woundering how you derive the minus sign into the equation m = -s´/s. In our textbooks m is defined by dividing the absolute value of s´ with the absolute value of s? thank´s!
MrHonordeath
The minus sign is important. If the magnification (m) is positive, then the image is in the same direction as the object, and if the magnification is negative then the image is inverted relative to the object.
This is the standard treatment of thin lenses and it contains numerous mistakes the best thing to do with this video is to ignore it.
Very helpful! Thank you!
Can anyone help me with this ? The image of an object formed on the screen by a convex lens has height a. By moving the lens towards the screen , it is found that there is a second lens position at which another image of height b is formed on the screen. Prove that the height of the object is (ab)^0.5 .
Why don't you use sign convention
Does the height of the object not affect the image created?
+RootDubz93 In some cases it could, but generally no. Note that these are just diagrams.
Thanks for these video! :))) It really help me a lot :D especially the formulas :) Thanks for the advantage :D :)
This really helped alot
very nice ....
what would happen if f was negative?
f is negative for a diverging lens. There are lots of examples in the playlists.
thank a lot
1:32 The human ruler.
very nice
amazing
Why we dont do sign conversion
We do use a sign convention when dealing with lenses.
Thankyou sir your lecture are great. I prepare for my IIT JAM exam with the help of your lecture.
Thanks
wtf
THANK YOU!!!!!
Thank Youuuuu !!!! :)
Equation of focal length is wrong..it is -ve in b/w
where is the next part
Are you looking for this video? Physics - Optics: Lenses (2 of 4) Converging Lens ua-cam.com/video/P5Y6rmPpGo4/v-deo.html
Why haven't we taken s as negative 100?
Since the object is in front of the lens, S should be positive.
Michel van Biezen
Taking that Into consideration shouldn't s' be negative then because it's behind the lens
No, by definition, s' is positive if it is behind the lens. (That is where you expect it to be if the rays travel through the lens.)
Sir what is R for??
R is the radius of curvature of the lens.
@@MichelvanBiezen thank you
Concave has -f ,while Convex has +f. Am I right?
That is correct.
You are great...but you need to go soooo much slower, so confusing, I had to repeat a lot... I for one like the bow tie
Yes, I tend to go a little too fast. (I am trying to slow down with the newer videos).
Great! thanks :)
I want a teacher with a bowtie too! 💃
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We have thousands of videos on physics (and continue to add more)
What's Hamilton canonical equation of this oscillator?
I don't see the connection to this video.
In all the optic problems, how can we tell whether the focus is negative or positive?
Look at these videos for your answer: PHYSICS 55.2 THE LENS EQUATION