So, to best summarize the lecture, visual acuity is the ability to best distinguish between two point. When two point make a 1-minute angle ( which is 1/60 angle) at node of eye(which is a point between 2/3 anterior and 1/3 posterior of eye) which places on the retina two point between them is a distance of 4.5 u will allow one cone to be stimulated instead of its adjacent cone. Otherwise stimulation of the cone with its adjacent cone will not help in distinguishing between two points. Cones themselves have a width of 2.5-2.8 u so that is the reason for 4.5 u distance which leaves one cone between the two stimulated cones. Looking at fovea, it has cones that are compressed due to depression and it's width is 2-2.2u which lessen the needed 4.5u so that causes better visual acuity, also one cone is concerned with one nerve fiber (unlike the rodes concerned with multiple nerve fibers) and lessened threshold for stimulation.
Sir, is there any relation between the visual acuity measured and the power of the lenses required by the person with that much visual acuity to see properly. That is to subtend the one minute angle at the nodal point of the eye.
Saw the msg late, i think. Yes, what kind of power of the lens is used for correction would depend on the visual acuity at that moment. Generally, the glasses (power) is used to take the vision to 6/6 (or, 6/9).
Sir even though cones are denser and thinner(i.e., 2-2.2micron)at fovea...To see two objects seperately we however need 4.5micron distance between 2 cells so like this if cones are thinner then more cones will be unresponsive na sir...
If thickness of a cone is 3 microns, then about 3.2 microns distance can be left between the two image points on the retina (so that they can be perceived as two points). But if the thickness of a cone is only 2 microns, then about 2.2 microns distance (lesser distance) can be left between the two image points on the retina. The issue is - one cone should be left unstimulated between the two points. Slenderer cones would mean, lesser distance can be left (so as to leave one cone unstimulated between the two points). Image points closer on the retina means the actual object points can be closer and still can be recognized as two separate points. It means a better visual acuity.
Retinal distance is not a fixed issue (like, compulsorily 4.5 microns have to be left between two image points). What is fixed is the fact that at least one cone should be left unstimulated between any two image points on the retina.
@@VivekSirsPhysiology Can you have it by thursday? I have an exam on friday and im having trouble understanding it but based on your recent videos, you have explained things clearly. Thank you!
This is called differential stimulation of cones. If images of two points fall on the adjacent cones, they will be excited with a same stimulus value. In that case, although the points may just be seen, the images will be blurred. There won't be a clear distinction between those two points.
'Seeing' and 'looking at' are two different things. We can see a wide range of external world, as most part of retina is sensitive. However, while looking at a specific object, we need the finer details of the object. So we look at the object in such a way that the light from the object falls on the fovea. This arrangement is for a narrow focus on a particular object at one time. It saves our cortex burdening with unnecessary load of information. Only those objects of particular concern or interest are focussed on the fovea. Others, we can just see, no details required.
@@Apratim98 Image will be blurred if it is not forming on the retina (forming either in front or behind the retina). Also, it has to the details of the image have to stimulate two separate cones, with one cone left unstimulated between the two points. If the image of the details is wider , so as to stimulate two adjacent cones simultaneously, again the image will be blurred.
Placing a lens in front of the eye causes and additional refraction of light rays. It has effect on the overall size of the object (object appears little smaller). However, there is no noticeable difference in the visual acuity. Normally, 1' angle corresponds to 4-4.5 micron distance on the retina. And thickness of a single foveal cone is not more than 3 microns. (The prerequisite for seeing two points separately is leaving at least one cone unstimulated between two point images.) Even if there is an additional refraction with lens placed in front of the eye, the light rays would still leave one cone unstimulated between the two image points on fovea.
Clarification....does this mean that the minimum separable distance is less than 1' for the fovea centralis region? Since the cones are thinner and more slender compared to other regions?
Yes. For the fovea centralis, the angle can be even smaller. This is why we turn our eyes and look at the object when we have to see its minute details. The image should be forming on the fovea.
VA=R 6/36 L 6/60. PH = R 6/36 L 6/60 AR = R -2.250/-1.00×42°. L -2.50/-0.250×1° sir Kya Mera low vission ka certificate ban sakta hai Kya Meri both eyes Mai ptosis hai aur CPEO( Cronical progressive external opthalmologia) hai aur Kearns saere syndrome hai aur diplopia bhi hai
If the gap between two photoreceptors is 4.5 microns,then there must be two photoreceptors should be unstimulated because a photoreceptor width is around 2.5-2.8 microns right.i don't understand this clearly sir
Whether it is 4.5 microns or 4 microns on the retina, and whether it is 2.5 microns or 3 microns width of a single photoreceptor; keep the numbers aside for the time being. And, just concentrate on the concept - AT LEAST one photoreceptor should be left unstimulated between the two point’s images on the retina. AT LEAST. It means, if more than 1 photoreceptor is left unstimulated between the two images, that will also be fine. So, if 4.5 microns distance is leaving two photoreceptors unstimulated between the images, it’s fine. The images will be perceived separately. The point is :- Those images should not form on the ADJACENT photoreceptors. And, this 4 to 4.5 microns distance ensures that adequately and easily.
Searching for this since long ! Thanks you tube! However you did not consider the distance between object and eye which plays important part in acuity!
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The best teacher of physiology he knows where we lack....we lack at the micron level of concepts n takes up those concepts 👌👌👌👌
Thanks for the comment..! Hope the channel is useful for Internal medicine students (or, in fact, to all streams)…, Regards,
Excellent!!
Sir has explained in a really clear way..❤️
So, to best summarize the lecture, visual acuity is the ability to best distinguish between two point. When two point make a 1-minute angle ( which is 1/60 angle) at node of eye(which is a point between 2/3 anterior and 1/3 posterior of eye) which places on the retina two point between them is a distance of 4.5 u will allow one cone to be stimulated instead of its adjacent cone. Otherwise stimulation of the cone with its adjacent cone will not help in distinguishing between two points. Cones themselves have a width of 2.5-2.8 u so that is the reason for 4.5 u distance which leaves one cone between the two stimulated cones.
Looking at fovea, it has cones that are compressed due to depression and it's width is 2-2.2u which lessen the needed 4.5u so that causes better visual acuity, also one cone is concerned with one nerve fiber (unlike the rodes concerned with multiple nerve fibers) and lessened threshold for stimulation.
Yes. Thanks for the summary…! Regards,
Just perfect💯!
God bless you, Sir!
The way you explained...hats off to u sir.....crystal clear
Thanks !
Very well explained, it makes me all cleared about visual acuity. Thank U Sir🙏🙏🙏🙏🙏🙏🙏
Thanks,
Regards,
Very nicely explained
Amazing explanation...😍
Thank you so much. This was so helpful. Please more videos
Sir please make more videos on eye
Nice video!!!👍
Its very helpful.. Thankiu so much sir👏
Hello youtube ...please give 3x option... good lecture
Sir, is there any relation between the visual acuity measured and the power of the lenses required by the person with that much visual acuity to see properly. That is to subtend the one minute angle at the nodal point of the eye.
Saw the msg late, i think. Yes, what kind of power of the lens is used for correction would depend on the visual acuity at that moment. Generally, the glasses (power) is used to take the vision to 6/6 (or, 6/9).
Sir even though cones are denser and thinner(i.e., 2-2.2micron)at fovea...To see two objects seperately we however need 4.5micron distance between 2 cells so like this if cones are thinner then more cones will be unresponsive na sir...
If thickness of a cone is 3 microns, then about 3.2 microns distance can be left between the two image points on the retina (so that they can be perceived as two points).
But if the thickness of a cone is only 2 microns, then about 2.2 microns distance (lesser distance) can be left between the two image points on the retina.
The issue is - one cone should be left unstimulated between the two points.
Slenderer cones would mean, lesser distance can be left (so as to leave one cone unstimulated between the two points). Image points closer on the retina means the actual object points can be closer and still can be recognized as two separate points. It means a better visual acuity.
Retinal distance is not a fixed issue (like, compulsorily 4.5 microns have to be left between two image points). What is fixed is the fact that at least one cone should be left unstimulated between any two image points on the retina.
@@VivekSirsPhysiology Thank you very much sir for taking ur valuable time in clearing my doubt🤩🤩🤩🤓🤓🤓
Sir what is vernier acuity and its purpose?
Is 4.5 microns is a bit that could come in exams?
can you make a video on countercurrent multiplication and countercurrent exchange in the kidney?
Ok. In the coming week.
Regards,
@@VivekSirsPhysiology Can you have it by thursday? I have an exam on friday and im having trouble understanding it but based on your recent videos, you have explained things clearly. Thank you!
@@dannyn7939
Yes . I'll try to make the video and upload it as soon as possible .
For distinction of 2 points, why is it required for atleast 1 cone to be unstimulated in between?
This is called differential stimulation of cones. If images of two points fall on the adjacent cones, they will be excited with a same stimulus value. In that case, although the points may just be seen, the images will be blurred. There won't be a clear distinction between those two points.
I got the whole think.. But values are totally different from as given in Guyton..
Sir why isnt all our retina have high acuity?
'Seeing' and 'looking at' are two different things. We can see a wide range of external world, as most part of retina is sensitive. However, while looking at a specific object, we need the finer details of the object. So we look at the object in such a way that the light from the object falls on the fovea.
This arrangement is for a narrow focus on a particular object at one time. It saves our cortex burdening with unnecessary load of information. Only those objects of particular concern or interest are focussed on the fovea. Others, we can just see, no details required.
@@VivekSirsPhysiology thank you sir i get it it now...
@@VivekSirsPhysiology Sir also if you would be kind enough to tell me why image is blurred if focus is not point sized on retina?
@@Apratim98 Image will be blurred if it is not forming on the retina (forming either in front or behind the retina). Also, it has to the details of the image have to stimulate two separate cones, with one cone left unstimulated between the two points. If the image of the details is wider , so as to stimulate two adjacent cones simultaneously, again the image will be blurred.
Sir does that mean only in foveal vision we see details.that is we have clear vision in narrow area only.then why done have huge visual field sir
what is the meaning of normal and subject?
Normal vision = 6/6
Subject is someone on whom the test is performed.
Can someone explain to me what happens to the lightrays that form a 1' angle when we put lens infront of the eye ?
Placing a lens in front of the eye causes and additional refraction of light rays. It has effect on the overall size of the object (object appears little smaller). However, there is no noticeable difference in the visual acuity. Normally, 1' angle corresponds to 4-4.5 micron distance on the retina. And thickness of a single foveal cone is not more than 3 microns. (The prerequisite for seeing two points separately is leaving at least one cone unstimulated between two point images.) Even if there is an additional refraction with lens placed in front of the eye, the light rays would still leave one cone unstimulated between the two image points on fovea.
Clarification....does this mean that the minimum separable distance is less than 1' for the fovea centralis region? Since the cones are thinner and more slender compared to other regions?
Yes. For the fovea centralis, the angle can be even smaller. This is why we turn our eyes and look at the object when we have to see its minute details. The image should be forming on the fovea.
VA=R 6/36 L 6/60. PH = R 6/36 L 6/60 AR = R -2.250/-1.00×42°. L -2.50/-0.250×1° sir Kya Mera low vission ka certificate ban sakta hai Kya Meri both eyes Mai ptosis hai aur CPEO( Cronical progressive external opthalmologia) hai aur Kearns saere syndrome hai aur diplopia bhi hai
Opthalmologist hee iska better jawaab de sakta hai. Kisi nazdeeki ophthalm se consult karna sahi hoga.
If the gap between two photoreceptors is 4.5 microns,then there must be two photoreceptors should be unstimulated because a photoreceptor width is around 2.5-2.8 microns right.i don't understand this clearly sir
Whether it is 4.5 microns or 4 microns on the retina, and whether it is 2.5 microns or 3 microns width of a single photoreceptor; keep the numbers aside for the time being. And, just concentrate on the concept - AT LEAST one photoreceptor should be left unstimulated between the two point’s images on the retina. AT LEAST.
It means, if more than 1 photoreceptor is left unstimulated between the two images, that will also be fine. So, if 4.5 microns distance is leaving two photoreceptors unstimulated between the images, it’s fine. The images will be perceived separately. The point is :- Those images should not form on the ADJACENT photoreceptors. And, this 4 to 4.5 microns distance ensures that adequately and easily.
@@VivekSirsPhysiology thank you for clarifying the doubt sir
Searching for this since long ! Thanks you tube! However you did not consider the distance between object and eye which plays important part in acuity!
Actually he did.. If distance between lens and the points changes...that will also change the angle between the points.
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