1. The approximate power on the vertical meridian for the Rx +3.00 -1.50 x 102 is ______. Give the power rounded to two decimal places, not to 1/8 diopter steps.
Vertical meridian is the power in the 90th meridian. First we need to subtract our axis(102 in this problem) from the meridian we are looking for(90 in this problem). This will be 
![\ F_{\theta} = [(\sin \alpha)^2 \times F_{cyl}] + F_{sph}](http://s.wordpress.com/latex.php?latex=%5C%20F_%7B%5Ctheta%7D%20%3D%20%20%5B%28%5Csin%20%5Calpha%29%5E2%20%5Ctimes%20F_%7Bcyl%7D%5D%20%2B%20F_%7Bsph%7D%20&bg=ffffff&fg=000000&s=2 "\ F_{\theta} = [(\sin \alpha)^2 \times F_{cyl}] + F_{sph}")
is the power in the meridian you are trying to find is the difference in the axis and the meridian you are trying to find
is the cylinder of the prescription
is the sphere power of the prescirption![\ F_{\theta} = [(sin \12)^2 \times (-1.50)] + 3.00](http://s.wordpress.com/latex.php?latex=%5C%20F_%7B%5Ctheta%7D%20%3D%20%20%5B%28sin%20%5C12%29%5E2%20%5Ctimes%20%28-1.50%29%5D%20%2B%203.00&bg=ffffff&fg=000000&s=2 "\ F_{\theta} = [(sin \12)^2 \times (-1.50)] + 3.00")
![\ F_{\theta} = [0.0432273\ \times\ -1.50] + 3.00](http://s.wordpress.com/latex.php?latex=%5C%20F_%7B%5Ctheta%7D%20%3D%20%5B0.0432273%5C%20%5Ctimes%5C%20%20-1.50%5D%20%2B%203.00&bg=ffffff&fg=000000&s=2 "\ F_{\theta} = [0.0432273\ \times\ -1.50] + 3.00")
So you have a total power of +2.94 D on the 90th meridian
2. If the Rx reads -10.25 -2.50 x 040, what is the approximate power on the horizontal meridian? Give the power rounded to two decimal places, not to 1/8 diopter steps.
Horizontal meridian is the power on the 180th meridian. In this case you can subtract 40 from 180 or 0 from 40 since it is either 0 or 180.
![\ F_{\theta} = [(\sin \40)^2 \times (-2.50)] + (-10.25)](http://s.wordpress.com/latex.php?latex=%5C%20F_%7B%5Ctheta%7D%20%3D%20%20%5B%28%5Csin%20%5C40%29%5E2%20%5Ctimes%20%28-2.50%29%5D%20%2B%20%28-10.25%29&bg=ffffff&fg=000000&s=2 "\ F_{\theta} = [(\sin \40)^2 \times (-2.50)] + (-10.25)")
![\ F_{\theta} = [0.413176 \times (-1.50)] + (-10.25)](http://s.wordpress.com/latex.php?latex=%5C%20F_%7B%5Ctheta%7D%20%3D%20%20%5B0.413176%20%5Ctimes%20%28-1.50%29%5D%20%2B%20%28-10.25%29&bg=ffffff&fg=000000&s=2 "\ F_{\theta} = [0.413176 \times (-1.50)] + (-10.25)")
![\ F_{\theta} = [-0.619764] + (-10.25)](http://s.wordpress.com/latex.php?latex=%5C%20F_%7B%5Ctheta%7D%20%3D%20%20%5B-0.619764%5D%20%2B%20%28-10.25%29&bg=ffffff&fg=000000&s=2 "\ F_{\theta} = [-0.619764] + (-10.25)")
Hi, I am a NC Licensed Optician that is sitting for the SC board in June, 2010 I was wondering do you have any problems on: with the answers
Calculate the “power” of the cylinder in an oblique meridian. I am reviewing now for the board and this board uses equal degree and will be using an Oblique Axis Power Chart to find the OMP.
I also was wondering if you have any problems and the answers on determine the meridian of highest absolute power.
I have been working on problems and just needed extra study of this.
Ginger