Agreed ... but I would argue that your description is the same thing as described in this video. It's ALWAYS about vergence of light ... whether we are describing converging light (as in this lecture) or coupled optical systems (like you are describing). I prefer the simpler explanation, and try to avoid discussing real/inverted images, far points, and terms like puntum remotum because they don't help someone learn retinoscopy. Appreciate the input.
Great illustrations. One correction, however, would be that the subtraction of working distance actually has nothing to do with the vergence of light entering the eye. Instead, it is subtracted because the neutral reflex is found when the practitioner's eye matches the punctum remotum of the neutralized eye. Essentially, the patient is an artificial myope until this working distance is corrected by removing 1.50D or 2.00D (or whatever the dioptric equivalent of your working distance is).
@MarkDavidKnight for a 66 cm working distance its subtract 1.50 DS (1/66cm =1/0.66m=1.50)
2mbilal 3 weeks ago
@2mbilal I see... Got it. Thanks for explaining. Appreciate it :)
MarkDavidKnight 20 hours ago
Hi Dr. Thanks for the lectures, but isn't working distance 2.50 and not 1.50? Thanks Dr
MarkDavidKnight 2 months ago
amazing!
saliha109 3 months ago
@fofocientos
Agreed ... but I would argue that your description is the same thing as described in this video. It's ALWAYS about vergence of light ... whether we are describing converging light (as in this lecture) or coupled optical systems (like you are describing). I prefer the simpler explanation, and try to avoid discussing real/inverted images, far points, and terms like puntum remotum because they don't help someone learn retinoscopy. Appreciate the input.
eyevideofan 5 months ago
Great illustrations. One correction, however, would be that the subtraction of working distance actually has nothing to do with the vergence of light entering the eye. Instead, it is subtracted because the neutral reflex is found when the practitioner's eye matches the punctum remotum of the neutralized eye. Essentially, the patient is an artificial myope until this working distance is corrected by removing 1.50D or 2.00D (or whatever the dioptric equivalent of your working distance is).
fofocientos 5 months ago in playlist ophthalmology
00:39 what is that word? throtter / feropter??? can anyone help, thanks!
dounoewhoim 6 months ago
@dounoewhoim phoropter
quickshift75 5 months ago
@dounoewhoim Actually, he said phoropter
MarkDavidKnight 20 hours ago
This is excellent! Such a good explanation :D
MeritCullen 6 months ago
Very informative
faisalndm 6 months ago
Very in formative
faisalndm 6 months ago
Superb ! a very good presentation !
anumeha24 8 months ago
Thanks a lot for a wonderful xplanation.
mercadotecnico 9 months ago
U ARE GIFTED WITH GREAT TEACHING SKILLS!! Made a medico's life so much easier!! Thank u!! :)
annegreen90 11 months ago 2
What a wonderful presentation. I am an "Engineer" and I always wondered how this was done -- EXACTLY. Thanks, Otis
otissumnerbrown 1 year ago
great video!!thanks alot...
SuperGhosttrain 1 year ago
Dr. Root, you are a genius. This movie is the best retinoscopy lesson ever given!
mhmshahi2002 1 year ago
Muito bom, pena que é em inglês. Gostaria de saber como traduzir?
Marcos
MSA35532651 1 year ago