ArrayofJaguars, That is a good question. 25 cm is the accepted distance a "typical" person with good eye sight can read comfortably. As people get older this distance increases. That is why the magnification of a microscope depends on who is looking through it.
Sir you lectures are super. They help me everyday in re-learning the physics that I have forgotten from my own school days. Thanks for taking the time and effort to upload these brilliant tutorials.
So for strained vision, he only calculated magnification of objective lens. To get to total magnification of the compund microscope for strained vision, you probably have to multiply Mo(strain) and Me(strain), which will be larger than X2.6.
The ray diagram correctly shows the ray entering the second lens (eyepiece) to be diverging (and not parallel) to the principle axis....it will therefore converge behind the focal point of the eyepiece, not at its focal point. Extrapolation will however still give a large (and negative) magnification factor, albeit not as high as would be expected if this ray really did converge at the eyepieces focal length.
Want to thank u from bottom of my heart for Ur help...u don't know what a great job u r doing...gained confidence in Phys just bcz of u😊😊:-) greetings from india
Sir, if i am not wrong for the eye piece the object is the image from the objective lens. That means that the object of the eye piece is infact an imaginary object but its on the left side of the eye piece. So, should I consider the distance of the eye piece object to be negative or positive? Like lets say a problem asks me to find the focal length of the eye piece and the problem provides the necessary lengths to find that without providing the correct sign. The problem asks me to use correct sign convention to find the eye piece focal length.
Basically, if you place an object exactly at the focal plane of the eye-piece lens, then you expect the formation of the image at infinity. Therefore, I wonder how one can expect the formation of the image inside the focal plane of the second lens! 🤔
@@MichelvanBiezen Sure, then how by placing the lens exactly at the focal point of the first lens as the objective (time 1:16) we can form a real image inside the focal point of lens 2 (eye-piece)? What I expect is to see the formation of the image at infinity. Would you please clarify this point?
There's something I'm missing: rewatching Lecture 7: Lens Combinations - Two Converging Lenses you tackle the problem in a completely different way. Here you don't draw the second line from the top of the object to the focus point of the first lens in order to find the Image of the first lens, why?
Why don't microscopes use a curved mirror instead of the objective lens? The main benefit of using a small curved mirror that I'd expect is reducing the colour dispersion or chromatic aberration.
Yes but my question was about microscopes. The only light-based microscopes I've seen rely on lenses but it seems like a mirror could replace the objective lens if it was designed to bounce light to the side instead of in front of it.
Oops, you were asking about microscopes. (I just gave a lecture about telescopes, so I was still thinking about telescopes. In the case of microscopes it is important to have the objective lens very close to the object you are viewing. That cannot be done with a mirror which would be in the back of the microscope.
Yeah the very short focal length would be weird for a curved mirror. bouncing the light away from the source would be difficult too. Unlike most telescope mirrors, the mirror would need to be a far off axis and not symmetrical at all. Thanks for the quick replies.
Michel if you got the opportunity, please put some videos on "magnetism in matter" thats also a chapter which is included for exam and i dont find any usefull videos on that. but i will never mind if you cant do that. thanks
Javad, Can you explain a little more what you mean with magnetism in matter? I would be glad to, but I want to understand better what you are asking. Thanks
+Michel van Biezen He probably meant to say, "Magnetism and Matter", which is a lesson taught in 12th grade CBSE physics. It's a lesson dealing with matter (ferromagnetic, paramagnrtic, etc) and it's interactions with magnetic fields. It touches on the topic of temporary and permanent magnets (retentivity and coercitvity) and the earth's magnetic field. (its' vertical and horizontal components). Finally, the concept of the Bohr's magneton is touched.
the lens equation used in this video is 1/f = 1/v +1/u {here v=image distance , u= object distance } whereas in some other textbooks they use 1/f = 1/v - 1/u , using the later one i got different value for object distance u= vf/f-v substituting this i m getting a different value for m1 = -(f1 + f2 - L)/f1 instead of what u got in this video... bit confused...
The standard format is with the "+" sign. I have actually never seen the equation with the "-" sign. Read the text around the equation to see why they use a negative sign.
Eye pieces are converging lenses with a very short focal lens They are typically two or 3 lenses placed together in a barrel to reduce distortion you will get from a single lens. However the combined effect is like that from a single converging lens.
Sir please help...where is the final image formed when the microscope is in normal adjustment...is it at infinity or near point?? Also,in some questions the position of final image is not given...so should I take the final image to be at infinity or at near point??... please help...
Good questions. The answer is the same as for a magnifying glass, which means how the user wants to use it. With relaxed eyes the image is allowed to "float" farther away, reducing the overall magnification. With strained eyes, the image can be "pulled" in closer to a reading distance (typically considered 25 cm) which will increase the magnification.
@@MichelvanBiezen Thanks so much sir....in some questions they have put normal adjustment when image is at near point... that's why I asked...so,if no details regarding position of final image for a compound microscope are given,shall I take it to be at infinity? Once again thank you sir...
ArrayofJaguars,
That is a good question.
25 cm is the accepted distance a "typical" person with good eye sight can read comfortably.
As people get older this distance increases.
That is why the magnification of a microscope depends on who is looking through it.
Sir you lectures are super. They help me everyday in re-learning the physics that I have forgotten from my own school days.
Thanks for taking the time and effort to upload these brilliant tutorials.
So for strained vision, he only calculated magnification of objective lens. To get to total magnification of the compund microscope for strained vision, you probably have to multiply Mo(strain) and Me(strain), which will be larger than X2.6.
The ray diagram correctly shows the ray entering the second lens (eyepiece) to be diverging (and not parallel) to the principle axis....it will therefore converge behind the focal point of the eyepiece, not at its focal point. Extrapolation will however still give a large (and negative) magnification factor, albeit not as high as would be expected if this ray really did converge at the eyepieces focal length.
Want to thank u from bottom of my heart for Ur help...u don't know what a great job u r doing...gained confidence in Phys just bcz of u😊😊:-) greetings from india
Sir ur lecture are best I am indian and I'm understanding ur lecture properly thanks a lot sir 🙏
Great. Welcome to the channel!
Thanks so very much for this explanation. It most certainly helped.
Glad it helped!
Sir, if i am not wrong for the eye piece the object is the image from the objective lens. That means that the object of the eye piece is infact an imaginary object but its on the left side of the eye piece. So, should I consider the distance of the eye piece object to be negative or positive? Like lets say a problem asks me to find the focal length of the eye piece and the problem provides the necessary lengths to find that without providing the correct sign. The problem asks me to use correct sign convention to find the eye piece focal length.
Basically, if you place an object exactly at the focal plane of the eye-piece lens, then you expect the formation of the image at infinity. Therefore, I wonder how one can expect the formation of the image inside the focal plane of the second lens! 🤔
If the first image is not placed inside the focal point of the eye piece, the final imagen will either be out of focus, or not visible.
@@MichelvanBiezen Sure, then how by placing the lens exactly at the focal point of the first lens as the objective (time 1:16) we can form a real image inside the focal point of lens 2 (eye-piece)? What I expect is to see the formation of the image at infinity. Would you please clarify this point?
There's something I'm missing: rewatching Lecture 7: Lens Combinations - Two Converging Lenses you tackle the problem in a completely different way. Here you don't draw the second line from the top of the object to the focus point of the first lens in order to find the Image of the first lens, why?
Because that is not how it works. Drawing ray diagrams for a two lens system can be complicated.
@@MichelvanBiezen so ray diagrams are useful just for simple 1 lens problems?
Yes, and for 2 lens problems, they are more difficult to implement.
thank you sir
may you have a nice day
In 2:42, you mention that this is a virtual but upside down image. Shouldn't it be a virtual, upright image?
No, all images seen through a telescope are upside-down.
great video, thanks.
Why don't microscopes use a curved mirror instead of the objective lens? The main benefit of using a small curved mirror that I'd expect is reducing the colour dispersion or chromatic aberration.
Most telescopes indeed use the reflecting mirror. They are called reflecting telescopes instead of refracting telescopes
Yes but my question was about microscopes. The only light-based microscopes I've seen rely on lenses but it seems like a mirror could replace the objective lens if it was designed to bounce light to the side instead of in front of it.
Oops, you were asking about microscopes. (I just gave a lecture about telescopes, so I was still thinking about telescopes. In the case of microscopes it is important to have the objective lens very close to the object you are viewing. That cannot be done with a mirror which would be in the back of the microscope.
Yeah the very short focal length would be weird for a curved mirror. bouncing the light away from the source would be difficult too. Unlike most telescope mirrors, the mirror would need to be a far off axis and not symmetrical at all. Thanks for the quick replies.
Thank you very much sir!
Fantastic lecture!
Michel if you got the opportunity, please put some videos on "magnetism in matter" thats also a chapter which is included for exam and i dont find any usefull videos on that. but i will never mind if you cant do that. thanks
Javad,
Can you explain a little more what you mean with magnetism in matter?
I would be glad to, but I want to understand better what you are asking.
Thanks
+Michel van Biezen
He probably meant to say, "Magnetism and Matter", which is a lesson taught in 12th grade CBSE physics. It's a lesson dealing with matter (ferromagnetic, paramagnrtic, etc) and it's interactions with magnetic fields. It touches on the topic of temporary and permanent magnets (retentivity and coercitvity) and the earth's magnetic field. (its' vertical and horizontal components). Finally, the concept of the Bohr's magneton is touched.
+Javad Hikmati I guess it's there.
Where is the next video in this series?
The link to the next video is below in the description box.
Thank you
the lens equation used in this video is 1/f = 1/v +1/u {here v=image distance , u= object distance } whereas in some other textbooks they use 1/f = 1/v - 1/u , using the later one i got different value for object distance u= vf/f-v substituting this i m getting a different value for m1 = -(f1 + f2 - L)/f1 instead of what u got in this video... bit confused...
The standard format is with the "+" sign. I have actually never seen the equation with the "-" sign. Read the text around the equation to see why they use a negative sign.
What kind of lense do they use for the eyepiece? It sure as hell isn't a biconvex lens like the one drawn.
Eye pieces are converging lenses with a very short focal lens
They are typically two or 3 lenses placed together in a barrel to reduce distortion you will get from a single lens.
However the combined effect is like that from a single converging lens.
Sir please help...where is the final image formed when the microscope is in normal adjustment...is it at infinity or near point??
Also,in some questions the position of final image is not given...so should I take the final image to be at infinity or at near point??... please help...
Good questions. The answer is the same as for a magnifying glass, which means how the user wants to use it. With relaxed eyes the image is allowed to "float" farther away, reducing the overall magnification. With strained eyes, the image can be "pulled" in closer to a reading distance (typically considered 25 cm) which will increase the magnification.
@@MichelvanBiezen Thanks so much sir....in some questions they have put normal adjustment when image is at near point... that's why I asked...so,if no details regarding position of final image for a compound microscope are given,shall I take it to be at infinity?
Once again thank you sir...
where does the 25cm in the first formula come from?
It's the limit that your eyes can see without being strained. It's an accepted value.
I can't understand shit lol I'm stupid
🤫😉💕
thank you
Thank you