Jackson cross cylinder 

Thank you You have interchanged the items. It is 0.5 for axis and 0.25 for power.

I was trying to explain that if the cylinder change (by using JCC) is greater than or equal to 0.5, then you change the value of the sphere. In the example shown, the original prescription was -2.00 / -1.00 x 180. So the spherical equivalent becomes -2.50. If we add a -0.50 (because of JCC) to the cylinder, we get -2.00 / -1.50 x 180. Now the spherical equivalent is -2.75. so we add half the cylinder change of opposite sign to the sphere - so since we added a cylinder of -0.50, (to -1.00 to make it -1.50), we should add +0.25 to the sphere - so adding this to the sphere of -2.00, we get -1.75 / -1.50 x 180. The spherical equivalent of this is -2.50, which is the same as our original prescription (prior to JCC testing)

Thank you. I make the presentation in power point along with the animations, and screen record the slide show on my laptop while simultaneously doing the voice recording 🙂

@prasannashetty2526 , In the example given in the video, the cylinder was rotated in the direction of the minus axis of the Jackson cross cylinder becoz the cylinder in the trial frame was a minus cylinder. If the cylinder in the trial frame was a plus cylinder, then it should be rotated towards the plus axis of the Jackson cross cylinder

Are you an undergraduate or a postgraduate?

@@ophthalmologypearls UG mam

Out of curiosity, what is your special interest in this test? It is not required for an undergraduate to know this.

@@ophthalmologypearls den ok ma'am..thank u mam