Cross Retinoscopy just in 4 Simple Steps
Вставка
- Опубліковано 15 вер 2024
- #retinoscopy #crossretinoscopy #optometry
Join us:
✓UA-cam: / smartoptometry-with-samir
✓Android App:
play.google.co...
✓Blogsite:
www.smartopto....
✓Telegram link: t.me/smartopto...
✓Facebook Group:
www.facebook.c...
Very nicely explained.
▪️Stay with us and share with your friends.
Thank you,
Samir Sutradhar.
Sir i think final prescreption power is [ +1.50/+1.00 # 90' ]
+2.5DS/-1.0DC X 180 and +1.50DS/+1.0 X 90 Both are same...
Watch Simple Transportation to understand this: ua-cam.com/video/ymT1aloNBxM/v-deo.html
Thank you,
Samir Sutradhar
Rspected sir
After Minus WD minimum power is sphare* and diffrence between two power is cylidrical power and Will take axis to low power sphare **
@@kaushalkaushal9250
This is to get cylinder power in minus form. In spectacle lens, cylinder power is incorporated in Minus power. That's why we choose maximum plus and minimum minus as spherical power because it will give cylinder power in minus form.
In this optical cross you consider horizontal meridian as Spherical meridian. You explained that we can take any meridian as Spherical meridian. So in this optical cross if we consider vertical meridian as Spherical meridian the final optical cross power will be +1.50/+1.00×90° right?
Yeah, You are right.
@@optometry-with-samir Thank you @smartoptometry.
@@rajeshtripura725
Stay with us and share with your friends
How to find the final power in refraction recept. R.E +1.00-2.50,130&L.E +1.00+2.25,10 please tell me
@@SathwikM-si6zy
Watch How to calculate Retinoscopic power: ua-cam.com/video/TMM4yGalEik/v-deo.htmlsi=sPZb_rjRlK73wSjK
For calculating cyndrical we need to subtract 180degree power from 90 degree??
Watch this video about How to calculate Retinoscopic power: ua-cam.com/video/TMM4yGalEik/v-deo.htmlsi=4Db3lZXjXYuUoUQm
sir myopia less than WD in projection stage of retinoscopy then what will be the movement ? with or against?
If Myopia less than working distance power then without working distance power in place movement will be with and after placing working distance power, movement will be against.
Vertical 4 neutralize h or horizontal 3 h pr cross me wrong kio kr diya 🤔
Good review.
▪️Stay with us and share with your friends.
Thank you,
Samir Sutradhar.
Why is the degree of working distance always subtracted from the value of the correction lens derived from the retinoscope device? Can I prove this with physical proof and a mathematical formula? I need an answer please 🙏
What this video: ua-cam.com/video/BX6W7V5BpwY/v-deo.htmlsi=YodHHstvq69o2l8L
How to find pesoduphakia using retinoscopy reflex
Purpose of IOL is to make eye emmetropic for distance.
-Reflex will be very bright (Close to neutralize).
-IOL Edge can be seen sometimes.
-Posterior Capsular Opacity (PCO) against bright Reflex.
-Due to tilt of IOL sometimes bright specific areas can be seen.
Thank you,
Samir Sutradhar.
Why did you subtract working distance from cylinder also????
Here, we used Spherical trial lens for both meridian, so working distance will be subtracted from both meridian.
This technique is call "Retinoscopy with two Spherical trial lens".
There are 3 techniques of doing Retinoscopy. Watch here: ua-cam.com/video/ymHGZVQc2fY/v-deo.html
Sir I have a doubt .
Horizontal meridian is -3.00 , vertical meridian is+2.00 then how to calculate the refractive error
📌If Spherical meridian is Vertical:
▪️Sph power is +2DS
📌If Cylinder meridian is Horizontal:
▪️Cyl Power is: -3 - (+2.0) or -3 -2 or -5DC
▪️Cylinder will be 90° to cylinder meridian so axis is 90°
📌 Final power is: +2DS/-5DC X 90°.
Don't focus on number(High or Low) or signs( plus or minus).
Write down your formula and practice with pen or paper. By thinking you can't solve optical cross
If you didn't subtract working distance power then, subtract working distance power from both meridian.
So, horizontal meridian is -4.5 (-3.0 - 1.5, if working distance is 67cm) and vertical meridian is +0.50
📌If Spherical meridian is Vertical:
▪️Sph power is +0.5DS
📌If Cylinder meridian is Horizontal:
▪️Cyl Power is: -4.5 - (+0.50) or -4.5 - 0.5 or -5.0
▪️Cylinder will be 90° to cylinder meridian so axis is 90°
📌 Final power is: +0.50DS/-5DC X 90°.
When working distance is 1.5 then fine prescription is +0.50/-4.50×90
@@optometry-with-samir bhai sphare me to aap working distance liye hi nahi ho 😂
If vertical meridian is 90'
Then i think final number is
+150sph/+100cyl 180'
Watch this video about Optical Cross to understand how to write power from Optical Cross: ua-cam.com/video/TBOReXX0JCU/v-deo.htmlsi=5I6nU3aAdYOmmMFF
Final number glt 😂
You are wrong...
Correct yourself if you think final prescription is wrong.....
If you have any doubt share here, I will help you to clear it ....