hi, i wasnt looking for all these names and numbers- can you do a video that explains how a magnifying glass actually makes the image look bigger...the object stays the same no matter how you look at it but a magnifying glass makes the image look bigger and i dont understand how thats possible- thanks
d.c. There videos on the various lenses and a video on the magnifying lens. Look at playlists: PHYSICS 55 and PHYSICS 59 They will answer your questions.
+Arindam Bortamuly If you place the object just inside the focal point f ~ s, then the image will appear near infinity and you'll be able to look through the magnifying glass with relaxed vision. But that will not give you the maximum magnification. (see the second half of the video)
I didn't understand why we used m = θ'/θ . I know the approximations. That's not where I'm confused. I haven't seen m = θ'/θ in any of your videos and I wonder the derivation of this equation and how we use such a proportion. I think I didn't understand the angular size and what it means.
@@MichelvanBiezen Okay sir. I watched another video about this and kind of understood it. What I understand is θ is for the height of image before magnifying lens and θ' is for after the lens. We are comparing the images' height before and after in the eye ball. And the distance (d) between macula and eye lens is constant that allows us to compare angles. Let's say h'/S' = θ (without glass) and h''/S'' = θ' (with glass), S' = S'' = d. So, h''/h' = θ'/θ in small angles, of course. I fell like 90% got it.
Image distance = s*f / (s - f) = 25 cm magnification = -s'/s = -1/4 Therefore the area of image = (1/4)^2 * area of object = 1/16 (8 cm^2) = (0.5 cm^2)
i have a question. i thought, in the case of convex lens, as the object is moved from the focus towards the lens, the size of the virtual image keeps on decreasing, the highest magnification being when the object is just near the focus, the lowest when the object is just near the lens. then in this example, how does the magnification increase when the object is moved away from the focus?
There are two different kinds of magnifications. The one you are referring to is the ratio of the height of the image to the height of the object. That ratio does indeed decrease as the object moves away from the focal point. However the magnification of a magnifying glass depends on how large the image LOOKS to the viewer and does must take into account the size of the image AND the distance to the image. That is why the ratio of the angles is used.
Yes, as odd as it seems, the magnification of a magnifying glass depends on where the object is placed. When the object is placed in such a way that the image is close the magnification will be greater (25/f) + 1 When the object is placed in such way that the image appears far away, the magnification will be less (25/f). The "25" can be a different number and depends on the person's eye sight.
I have a question; given that we use a magnifying glass, how close to the lens should the eye be placed to get the maximum magnification but still see a clear image?
hello. can you explain why the magnification of a magnifying glass is f/25. I dont understand why theres a 25 at the bottom. couldn't the image be 30 cm away and the equation would be f/30? it just seems like its always going to be f/25 the way it was said in the video. thank you.
Master Chief, That is a very good question. It turns out that the magnification of a magnifying glass in part depends on the viewer as well. For most people the near point can be as close as 25 cm, and thus this is the typical value used for that equation. For others (like myself) who are becoming far sighted with age, the near point is larger and thus for a person like me the magnification would be more like f/40 or f/50. But 25 cm is the standard value used.
Um, I dot quiet understand on the second part of the video your said we will look through a strained vision , when why we still assume s' is still 25? since N=25 is relax vision
+JUnnnNnzzZ With relaxed vision, you use the same equation for the magnification: m = [(s'-f)/s'f] * (25/1) and let s' become infinite. Then m = (1/f) * (25/1) = 25/f
@@MichelvanBiezen Ok sir, however in the first exemple the distance between the lens and the object is not the same as the distance between the lens and the image (you said it will be formed in infinity) and we aplied the formula teta' / teta. What I am missing? P.S.: Sir, I really thank you for all those videos that you posted on youtube. It is really helping me on my Physics subjects. Keep the good work!
In the case that the image is at "infinity", the image will also be "infinitely" high. And thus the only way to indicate the magnification, is to use the ratios of angles to indicate the magnification.
Thank you Sir for your lectures, your methods are very easy to understand yet comprehensive. can u please explain the significance of the equation on the top right ( the one that goes like this: 1/f= 1/s + 1/s') please.
Mohamed Ayoob Sure. All 3 equations are the same. f is the focal length s is the distance from the lens to the object s' is the distance from the lens to the image
@@MichelvanBiezen Thank you! I truly appreciate all your videos. Your knowledge base is impressive. I'm delving deep into how my telescope actually works so these are very helpful.
Hi, first of all, thanks for the video! Second. I'd like to ask why it is that you can assume for small angles that tan theta = +/- theta? Looks like I missed out on something there
Benjamin, Check it out on a calculator. When you take the tan of a small angle (in radians) is becomes the same as the angle. In the limit as the angle approaches zero, it is the same.
Hi there I love the this video it has helped me to have a better understanding in this topic. BUT.. i have a confusion, isn't lens the formula (1/F)=(1/V)-(1/U). you seem to have done this (1/F)=(1/V)+(1/U) where U=S & V=S1 please correct me if i am wrong. thank you
+Varun Khandka I am curious as to why you think that the lens equation is not (1/f) = (1/s) + (1/s').This is a very standard equation. It could be that the book that you use has the sign of s' defined differently. Go ahead and check this. I can assure you that the equations I used are correct as defined by many text books.
+Michel van Biezen look at 6:02 when you solved the question S came out to be a positive value that implies that the object must be to the right side of the lens (that is: behind the eye), which is not possible. That is why i was wondering about the lens equation.
+Varun Khandka With a magnifying glass both the object and the image are on the left side (the front side of the lens). That means that the object distance is positive and the image distance is negative.
Thanks for the comment. I am glad it helped.
Your vids are a blessing, thank you!
Thank you so much, sir. Your video made me clear of what I was confused.
I needed this. Thanks so much for making this simple and easy to understand.
Glad it was helpful and that you found our videos.
Your videos are very helpful, Sir. Thank you so much
Glad to hear that
It would be great if you give us tutorial how head up display (HUD) works.
Thank you so much! Amazing video, helped me a lot.
thanks alot
if f be the focal length of the objective and f1 that of the eye piece then magnifying power is given by
hi, i wasnt looking for all these names and numbers- can you do a video that explains how a magnifying glass actually makes the image look bigger...the object stays the same no matter how you look at it but a magnifying glass makes the image look bigger and i dont understand how thats possible- thanks
d.c.
There videos on the various lenses and a video on the magnifying lens.
Look at playlists: PHYSICS 55 and PHYSICS 59
They will answer your questions.
Great explanation ! Thanks , it was very helpful.
For the sake of curiosity ! why did you choose 25 cm the distance to the object ?
It is the standard distance used in these types of problem. It is assumed that a good eye can easily see object clearly at a distance of 25 cm.
@@MichelvanBiezen " It is assumed ?" Thanks !
It is the standard assumption.
Sir why do we take angel to be very small so that its tangent is same as it is
why f=s?
Is it because to view an object clearly or to be in focus,
distance of object has to be equal to focal length?
+Arindam Bortamuly
If you place the object just inside the focal point f ~ s, then the image will appear near infinity and you'll be able to look through the magnifying glass with relaxed vision. But that will not give you the maximum magnification. (see the second half of the video)
+Michel van Biezen
Thank you so much Professor!
I didn't understand why we used m = θ'/θ . I know the approximations. That's not where I'm confused. I haven't seen m = θ'/θ in any of your videos and I wonder the derivation of this equation and how we use such a proportion. I think I didn't understand the angular size and what it means.
It is strictly based on geometry. For small angels if the image subtends twice the angle as the object it will look twice as big.
@@MichelvanBiezen Okay sir. I watched another video about this and kind of understood it. What I understand is θ is for the height of image before magnifying lens and θ' is for after the lens. We are comparing the images' height before and after in the eye ball. And the distance (d) between macula and eye lens is constant that allows us to compare angles. Let's say h'/S' = θ (without glass) and h''/S'' = θ' (with glass), S' = S'' = d. So, h''/h' = θ'/θ in small angles, of course. I fell like 90% got it.
Great!
A lens of focal length 20 cm has an object of area 8.0 cm2 placed 100 cm in front of it. The area of the image is
Image distance = s*f / (s - f) = 25 cm magnification = -s'/s = -1/4 Therefore the area of image = (1/4)^2 * area of object = 1/16 (8 cm^2) = (0.5 cm^2)
i have a question. i thought, in the case of convex lens, as the object is moved from the focus towards the lens, the size of the virtual image keeps on decreasing, the highest magnification being when the object is just near the focus, the lowest when the object is just near the lens. then in this example, how does the magnification increase when the object is moved away from the focus?
There are two different kinds of magnifications. The one you are referring to is the ratio of the height of the image to the height of the object. That ratio does indeed decrease as the object moves away from the focal point. However the magnification of a magnifying glass depends on how large the image LOOKS to the viewer and does must take into account the size of the image AND the distance to the image. That is why the ratio of the angles is used.
@@MichelvanBiezen ive been wondering about this my entire life, your explanation is so elegant, thank you good sir i can now sleep peacefully
What is the magnifying power of this magnifying glass if we vary the object distance .... does it changes if the focal length is fixed?
Yes, as odd as it seems, the magnification of a magnifying glass depends on where the object is placed. When the object is placed in such a way that the image is close the magnification will be greater (25/f) + 1 When the object is placed in such way that the image appears far away, the magnification will be less (25/f). The "25" can be a different number and depends on the person's eye sight.
I have a question; given that we use a magnifying glass, how close to the lens should the eye be placed to get the maximum magnification but still see a clear image?
sorry, not good. how does the light travel through the lens and eye?
That was not the purpose of this video. That is covered in other videos.
That was so helpful!! Thank you :)
hello. can you explain why the magnification of a magnifying glass is f/25. I dont understand why theres a 25 at the bottom. couldn't the image be 30 cm away and the equation would be f/30? it just seems like its always going to be f/25 the way it was said in the video. thank you.
Master Chief,
That is a very good question.
It turns out that the magnification of a magnifying glass in part depends on the viewer as well.
For most people the near point can be as close as 25 cm, and thus this is the typical value used for that equation.
For others (like myself) who are becoming far sighted with age, the near point is larger and thus for a person like me the magnification would be more like f/40 or f/50.
But 25 cm is the standard value used.
thank you for your response!
+Michel van Biezen
I had this question in my mind.
Very nice explanation!
Thanks! Nice video!
Um, I dot quiet understand on the second part of the video your said we will look through a strained vision , when why we still assume s' is still 25? since N=25 is relax vision
+JUnnnNnzzZ
With relaxed vision, you use the same equation for the magnification:
m = [(s'-f)/s'f] * (25/1) and let s' become infinite.
Then
m = (1/f) * (25/1) = 25/f
thank you so much for the promptly responds!
Sir, assuming that magnification is the ratio between the height of those 2 figures, you can't express it by teta' / teta.
Why did you do that?
You can if the distance between the lens and the object is the same as the distance between the lens and the image.
@@MichelvanBiezen Ok sir, however in the first exemple the distance between the lens and the object is not the same as the distance between the lens and the image (you said it will be formed in infinity) and we aplied the formula teta' / teta.
What I am missing?
P.S.: Sir, I really thank you for all those videos that you posted on youtube. It is really helping me on my Physics subjects. Keep the good work!
In the case that the image is at "infinity", the image will also be "infinitely" high. And thus the only way to indicate the magnification, is to use the ratios of angles to indicate the magnification.
I cannot see the board well.
Yes, the lighting was not as good with our older videos. (until we figured out what we were doing).
Thank you Sir for your lectures, your methods are very easy to understand yet comprehensive. can u please explain the significance of the equation on the top right ( the one that goes like this: 1/f= 1/s + 1/s') please.
Mohamed Ayoob
Sure. All 3 equations are the same.
f is the focal length
s is the distance from the lens to the object
s' is the distance from the lens to the image
thank you very much
Where are 17-20?
They will be posted over the next several weeks.
@@MichelvanBiezen Thank you! I truly appreciate all your videos. Your knowledge base is impressive. I'm delving deep into how my telescope actually works so these are very helpful.
Thanks, i like it.
I'm glad you like it
Good
Thanks 🙂
Hi, first of all, thanks for the video! Second. I'd like to ask why it is that you can assume for small angles that tan theta = +/- theta? Looks like I missed out on something there
Benjamin,
Check it out on a calculator. When you take the tan of a small angle (in radians) is becomes the same as the angle.
In the limit as the angle approaches zero, it is the same.
thanks nigga
lmao
Hi there I love the this video it has helped me to have a better understanding in this topic. BUT..
i have a confusion, isn't lens the formula (1/F)=(1/V)-(1/U).
you seem to have done this (1/F)=(1/V)+(1/U)
where U=S & V=S1
please correct me if i am wrong.
thank you
+Varun Khandka I am curious as to why you think that the lens equation is not (1/f) = (1/s) + (1/s').This is a very standard equation. It could be that the book that you use has the sign of s' defined differently. Go ahead and check this. I can assure you that the equations I used are correct as defined by many text books.
+Michel van Biezen look at 6:02 when you solved the question S came out to be a positive value that implies that the object must be to the right side of the lens (that is: behind the eye), which is not possible. That is why i was wondering about the lens equation.
+Varun Khandka
With a magnifying glass both the object and the image are on the left side (the front side of the lens). That means that the object distance is positive and the image distance is negative.
is it just me or he sounds like gru from despicable me anyways great lectures
It's not just you. Many people made that observation.
You sound like a Brooklyn-grown Steve Jobs! Haha :) Thanks so much for your videos :D
Pls answer