Cross Retinoscopy just in 4 Simple Steps

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Жыл бұрын

#retinoscopy #crossretinoscopy #optometry
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Пікірлер: 34

@swapankumarmallick3468 Жыл бұрын

Very nicely explained.

@optometry-with-samir Жыл бұрын

▪️Stay with us and share with your friends. Thank you, Samir Sutradhar.

@otiebrown9999 Жыл бұрын

Good review.

@optometry-with-samir Жыл бұрын

▪️Stay with us and share with your friends. Thank you, Samir Sutradhar.

@adnannoor2235 6 ай бұрын

sir myopia less than WD in projection stage of retinoscopy then what will be the movement ? with or against?

@optometry-with-samir 6 ай бұрын

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.

@user-oz8vm3ve3e 3 ай бұрын

For calculating cyndrical we need to subtract 180degree power from 90 degree??

@optometry-with-samir 3 ай бұрын

Watch this video about How to calculate Retinoscopic power: kzfaq.info/get/bejne/irN9Z9x4xtKpmZ8.htmlsi=4Db3lZXjXYuUoUQm

@kaushalkaushal9250 Жыл бұрын

Sir i think final prescreption power is [ +1.50/+1.00 # 90' ]

@optometry-with-samir Жыл бұрын

+2.5DS/-1.0DC X 180 and +1.50DS/+1.0 X 90 Both are same... Watch Simple Transportation to understand this: kzfaq.info/get/bejne/r9OEZMSd1LSmqIE.html Thank you, Samir Sutradhar

@kaushalkaushal9250 Жыл бұрын

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 **

@optometry-with-samir Жыл бұрын

@@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.

@iqra_life5494 Ай бұрын

Vertical 4 neutralize h or horizontal 3 h pr cross me wrong kio kr diya 🤔

@vaishaliganesan3261 Жыл бұрын

How to find pesoduphakia using retinoscopy reflex

@optometry-with-samir Жыл бұрын

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.

@rajeshtripura725 4 ай бұрын

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?

@optometry-with-samir 4 ай бұрын

Yeah, You are right.

@rajeshtripura725 4 ай бұрын

@@optometry-with-samir Thank you @smartoptometry.

@optometry-with-samir 4 ай бұрын

@@rajeshtripura725 Stay with us and share with your friends

@conanr.1138 9 ай бұрын

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 🙏

@optometry-with-samir 9 ай бұрын

What this video: kzfaq.info/get/bejne/eL5mipqHmqjUp40.htmlsi=YodHHstvq69o2l8L

@batsonplexus0013 Жыл бұрын

Why did you subtract working distance from cylinder also????

@optometry-with-samir Жыл бұрын

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: kzfaq.info/get/bejne/r9N4er2HtsmWlo0.html

@Kusuma427 Жыл бұрын

Sir I have a doubt . Horizontal meridian is -3.00 , vertical meridian is+2.00 then how to calculate the refractive error

@optometry-with-samir Жыл бұрын

📌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°.

@optometry-with-samir Жыл бұрын

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

@optometry-with-samir Жыл бұрын

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°.

@mtfactech6053 Жыл бұрын

When working distance is 1.5 then fine prescription is +0.50/-4.50×90

@mtfactech6053 Жыл бұрын

@@optometry-with-samir bhai sphare me to aap working distance liye hi nahi ho 😂

@KumarSD2415 8 ай бұрын

If vertical meridian is 90' Then i think final number is +150sph/+100cyl 180'

@optometry-with-samir 8 ай бұрын

Watch this video about Optical Cross to understand how to write power from Optical Cross: kzfaq.info/get/bejne/iqh_hciJvZauc4k.htmlsi=5I6nU3aAdYOmmMFF