Thin lens equation and problem solving | Geometric optics | Physics | Khan Academy
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- Опубликовано: 5 июл 2014
- Some examples of using the thin lens equation. Created by David SantoPietro.
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how one man can so clearly explain in 12 minutes what my professor fails to explain in 2 hours i will never understand
I think that they have a very deep knowledge of the subject
I think you do not pay attention in the classroom.
You have one sole aim of knowing this
Nobody forced you
U think that's why u understand better on RUclips and the video is rehearsed
You save so many lives everyday!
Not even 20 seconds in the video and David has already figured out my dilemma in this lesson---"When are these positives or negatives?". Literally made me excited that I was watching the correct video.
Fr
Sal, please put David's videos into the main list with physics and math videos, especially those about waves. These are the best lessons I've ever heard.
yeah no offense to Sal, but he is hard to listen to with all that stuttering... this guy is so smooth
anyone with exams tomorrow trying to study on youtube
me_irl
Yes
Yes
Present
Physics exam tmr, grade 7 haha life is tough
I was so depressed not understanding this for weeks and you just solved it out in 10 mins...
Crazy shit being taught in Grade 10 nowadays
lol
Pitching Wedge I am learning this in 8th grade
fr fuk this shit bro
Watching these the day before the exam like a crazed hobo frantically searching for cigarette butts on the street to get his kick
This man has likely single handedly boosted the worlds GPA
Yes, i think sal must add these videos in the playlist, it would be very convinient
This is remarkably articulated. Thank you.
This is the absolute BEST video I've watched for thin lens, thank you so much
Best Khan instructor I've watched, thank you for the help.
THANK YOU.
I never quite got around to learning this. Now I'm studying for the PGRE and optics is like my weakest area.
upload more and more lectures on every small or big topic☺ your doing greaaat work🙋🙋
youre the hero i needed since last week
I was so lost, thanks for the help!
This all makes so much sense.
After crying for 3 hours while looking at confusing lines, you have saved my grade in 12 minutes. Thank you!
im not even studdying for the mcat but this summed up 2 weeks of lecture in 10 minutes you are a fucking boss thanks
TRUTH BE TOLD why is this on the MCAT?
very great amazing spectacular video !! the dark bg and neon colors rly do help focus my attention 2 the screen, justa really overall great video in every aspect :)
Amazing! Thank you!
this is on my exam tomorrow and my teacher didn’t even teach this 😭 gr 10 btw
same here!
@@arolemaprarath6615 im in canada tho, idk about the other person
Grade 10 optics squad :)
lol same grade 10 gang !
Great examples!
Thank you so much for this explanation!
I usually don't have any problems with physics, but this issue was just too confusing. I have a book that mentions how to determine the signs, but it just made matters worse. I'd take my semi-understanding and attempt to answer an equation with it, and I just keep getting it wrong. This video makes it so amazingly simple. Again, thank you so much, you've been a great help.
Your videos are great, thank you
this is actually wow
You've literally saved my ass, it all makes sense now
Dude you are a life saver wow
nicely present. thanks
i like this guy more than the other one. he just seems less boring but helpful none the less
Thanks you for this video it help me very much 😍
Love this guy
thanks that helped
Good!!
The app you uses please it will help me make exersices and solve things
Thank you sooo so much!!
I've got a really shitty optics teacher that doesn't explain shit and this video really helped me understand thin lenses! thanks so much!
when i saw 6 instead of 12.....you had me in the first half not gone lie
Thank u 👍
excellent
BRAVO!!!!!!!
Which softeare do you use sir !
thanks
good lecture
Learning this in grade 7, some really crazy shit
please add a option in this app to ask questions
would b better if u could explain sign convention rules der like d graphs...stuff formed on d right side is +ve and stuff on d left side -ve and so on...
What about calculating strength of lens?
whether mirror and lens equations are same? for lens it should be (1/f = 1/v - 1/u) right?
this saved me
These comments are so helpful
AUEEESOMMMMEEEEEEE!
What about when object distance is less than the focal length for a convex lens? Doesn't that make S' negative, but it's on the opposite side of the lens than the object?
may God bless you
I was wondering if a very wide lens is used or the widest can simulate some special effects, like Interstellar movie from CN
THANK YOU! I am curious though. How do you know if an image is real or virtual?
If it is real, it can be focused onto a screen.
Virtual images can only be seen by eye and is on the same side as the object on the ray diagram
on one of the practice problems amcas gave the equation /0 + 1/i = s and asked to solve for the strength of the person's eye. Looking back it was probably a plug and chug especially since they gave the equation but it was the first time I have seen this equation. I'm only familiar with the solving for focal length and magnification. Are there any videos on lens strength anywhere?
10 months after, sorry.
But, I think by "strength " it meant using the lens power equation : P = 1/F where F is in meters.
Isn't the sign of the focal lengths opposite? + for concave and - for convex
for a concave mirror: f = +0.5R
for a convex mirror: f = -0.5 R
MintChocChip100 this is a lens
Hi! Thank you, Khan videos have helped me a lot! I just wanted to point out that the description of the lens use at 5:17 is not quite accurate in terms of the eye. That would be a _real image_, which you would not view in that manner (or at all). A real image is one that is actually formed on a surface, which could only be viewed from behind like that if there was, say, a plane of frosted glass or a piece of translucent paper. Real images are generally formed on film, sensors, retinas, etc. or viewed anew with your own eye forming _another_ image. In other words, a flat surface must be present to realize a real image, and it would probably be viewed from off axis.
Yes, positive diopter lenses can be used in that schematic layout for corrective purposes, but that is to partially converge light (and not form an image yet) before your eye lens can form a real image on your retina (which would be the case for far-sightedness -- nearsightedness would require a negative diopter lens).
Sorry, don't mean to be a stickler!
Oh. except if the image is inside the focal length, in which case the image would be upright and virtual. Magnifying glass style. Then yes, your eye would be there, but the rays would not be converging.
@@dskim24 dbi
U r so clutch
Does it matter where your eye is located? Does the image stay in focus whether your eye is 10cm away, 50cm away etc?
I'm trying to determine if there is a good position to place your eye so that the object is in focus or if focus matters? Referring to this product: www.shapeways.com/product/67MLBFUU4/vr-one?li=search-results-1&optionId=57570022
khan academy is doing god's work
I love your voice
isnt the image of concave lens always on the same side as the object? meaning the image distance will be negative??????
Yes
Henry Shingalili that's what I thought
Henry Shingalili no when object is placed in between optic centre and f2
love you
can you explain why the image is not inverted? im really confused @__@
we are taught that distance in front of the lens is negative and behind the lens is positive
Can someone answer this plz.. so, if the distance of image is positive= the image is inverted and real..?
If the distance of image is negative= the image is not inverted and virtual...?
+Syazlina Aasim in case of conves lens if distance of image is positive then image must be real(because light are actually meeting ) and inverted (because image is form below the principle axes ) .In case of concave lens if image is taken as negative thats mean image is real and inverted . hope you got.
you need a pay raise
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 .
Thought you had to convert to meters at the end when using magnification equation?
magnification is unitless. the top and bottom units cancel out if they're the same
how did u get 6 in 1/8-1/24
3 years late but here's your answer.
Common denominator:
1/di=1/do-1/f=f/f*do-do/f*do=f-do/f*do
Solve for di:
di=f*do/do-f=-8*24/24+8=-2*2*2*2*2*2*3/2*2*2*2*2=-6
Bro he probably already graduated 😂😂
Is it always positive, or negative??
I'm eight and what I am doing here
Now your 10 yo right?
@@yotamrandria8600 correct 🤣🤣
Now he’s 11
You're 6
Now ur 11
This is my University.
10:15 --- when we subtract we get -(1/32) not -(1/6)
your maths is beyond science
5:24 its SUPERMAN's EYE
Soooo how do you figure out the heights of the object and image......?
Borat Sagdiyev magnification
In convex lens, where did you get the 6cm? Sorry, I just didn't processed it correctly because i think you jump that part. Please answer 🥺
At 10:15, we can simplify 1/-8cm - 1/24 cm into -1/6cm by finding the least common multiple of the denominators (aka multiply the top and bottom of one of the fractions so that they will add easily) :
First, recognize that -1/8 is equivalent to -3/24. (multiply the numerator and denominator by 3).
With the same denominator, -3/24 and -1/24 can be added to become -4/24.
Simplifying -4/24, we get -1/6.
Why some other videos concace is positive and convex is negative. So which is the correct
For convex, it can be positive or negative depending on object distance from the lens.
Only when object distance is less than the focal length does the image distance become negative for convex lens.
What app did you use to make those doodles? thanks alot
I wonder about it too
W
This mans has better hand writing on his computer than I have with a pencil
same bro, idk I have to be happy for him or sad for me ) : (
can some1 pls give me a detailed solution/explanation on how he got rid of both the numerators (1) in 10:32
because the value of di is the reciprocal (a fraction flipped upside down) of -1/6, so if you flip this fraction, it is -6/1 which is -6. I hope thats more clear?
@@papapanda15 ah, yes! thank you for taking the time to explain ~~
we have been taught that object distance should always be negative?
that is when you're using cartesian sign conventions, in that case the lens formula becomes 1/f = 1/v - 1/u.
you should use the formula which your teacher has suggested you to avoid confusion.
How did this guy get 1/-6 from the equation like what did he do???
lmao i swear hes wrong the answer for di is -12
How did you get from 1/-8cm - 1/24 to -1/6cm ?
Madeleine Ezard Because of math. -1/8 - 1/24 = -3/24 - 1/24 = -4/24 = -1/6 = -6 cm.
+scuzum2u does that make sense? Scuzum, could you please pretty please explain again? :)
+Noo Rii Not sure if it is too late, did you figure it out? I assume you're not getting this because you want to avoid using a calculator, which you do not need. First find common denominators to change -1/8 - 1/24 by multiplying by 3/3 to both. Then it becomes -3/24 - 1/24 which is -4/24, which reduces to -1/6. But you have to take the reciprocal in order to find your f, di, or do.
+Sunbir Gill Kind Sir, thank you for your answer, I need to rewatch the video :) Maybe it will make sense with your explanation. Again thanks so much :)
scuzum2u lmaoooo I'm so retarded thank you
Interesting how he can't explain the Algebra II concept of rotational integration, aka rotational volumetric integration about the center. This should be continued. He probably knows it now lol
how the heck do you get -1/6?
how does he get -6
ayy, that's a pretty confusing video you got yourself there
I think you made a mistake regarding the sign of magnification and it indicating inverted/non-inverted image.
Sabah Khan ikr i thought if the magnification is (-) then its inverted and when the image is (+) its upright and erect
+ve or -ve also simplifies directions. What I have learnt is if anything two of these u, v and f have same signs then they are on the same side of the mirror. But that's not what he is saying.
If say u and v are both positive then how can they be on different sides..
THAT'S WRONG!!!!!!
how did you get 1/6th? -8-24 =32? right? or are we subtracting fractions? someone please explain!
We are subtracting fractions in this problem. Now in order to subtract fractions you must find common denominators. In this case 1/-8 and -1/24, the common denominator is 24, so multiply the numerator and denominator of 1/-8 by -3 to get -3/24, now you got a common denominator of 24, so you can subtract; -3/24 and -1/24 to get -4/24 you can now simplify this fraction by dividing the numerator and denominator by 4 to get -1/6. I too was confused for a bit because he skipped this step in the video.
how did you go from -8cm-24cm equals 6 😭
denominator has to be equal for adding fractions. (-1/8) x (3/3) = -3/24. (-3/24) - (1/24) = -4/24 = -1/6
Not good. The issue of this video is how the image is formed. It is missing information.
I'm having a total dead brain moment and cannot figure out how he got positive 1/6 from (-1/8) - (1/24), I keep getting 1/12, anyone care to explain?
When you're adding or subtracting fractions, you need common denominators. -1/8 becomes -3/24, so now you can subtract: (-3/24) - (1/24) = -4/24 (reduced form is -1/6). It's negative, not positive. Hope this helps!
okay but how the fuck do you get '-6' from '-8 - 24' ?
Where is Sal, dammit?