Hand Neutralization.pptx

Hand Neutralization.pptx

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  Hand Neutralization Ms. KhushiKansal Assistant Professor Swami Vivekanand Subharti University
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  INTRODUCTION  The powerof an unknown lens can be determined by neutralizing it with another lens of known power.  Neutralization is based on the fact that if you look at an object through a convex or concave lens and move the lens from side to side (right and left or up and down), the image that you see through the lens will also move. If the lens has no power - or has been neutralized by placing a lens of equal power but opposite sign against the unknown lens - then there is no image movement
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  MOVEMENTS  A pluslens moves the image in the opposite direction to the lens:  If you move the lens to the right, the image will move to the left.  If you move the lens up, the image will move down, and vice versa.
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  MOVEMENTS  A minuslens moves the image in the same direction to the lens:  If you move the lens to the right, the image moves right.  If you move the lens up, the image moves up, etc.
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  MOVEMENTS  A sphero-cylindricallens causes scissoring movements on rotation:  If you rotate the lens about its optical centre the image of a cross will appear to scissor.  The principle meridian are determined by marking the orientation at which the cross lines are perpendicular.
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   A lenswith no power causes no movement  Regardless of lens movement, the image remains stationary.
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  STEPS IN NEUTRALIZATION 1.Draw a cross: On a piece of paper draw a cross with lines perpendicular to each other and at least 15cm long. 2. Determine principle meridian: Look at the cross through the lens - the further away the cross is from you during this task.  Rotate the lens to determine if their is astigmatism by observing for scissoring movements of the cross image. If there are no scissoring movements, the lens is spherical (there are no principle meridian).  If scissoring movements are present, then rotate the lens such that the cross lines of the image are exactly perpendicular. This is the orientation of the principle meridian of the lens. You can now mark the principle meridian on the lens with spots using a felt-tip pen.  Keep the lens in this orientation for the remainder of the task.
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  STEPS IN NEUTRALIZATION 3.Determine the optical centre: Now move the lens up-and-down or side-to-side to ensure that the lines of the cross image exactly overlap with the lines of the object cross.  Mark the optical centre with a felt tip pen.  Neutralization of the lens should occur at the optical centre. 4. Neutralize each meridian: Holding the lens in orientation with its principle meridian horizontal and vertical as in step 3, move the lens from side to side.  If there is with movement, the lens is minus along this meridian, and you require a plus lens to neutralize. If there is against movement, the lens is plus along this meridian and you require a minus lens to neutralize.  Using a bracketing technique, find the lens that achieves neutralization along this meridian (i.e. no movement) and remember that the power of the unknown lens is the same size but opposite in sign to the neutralizing lens.  Once the horizontal meridian is neutralized, you can move the lens up and down to neutralize the vertical meridian or vice versa.
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  STEPS IN NEUTRALIZATION 5.Draw a power cross: The power of the unknown lens along the meridian tested is the same as the power of the neutralizing lens but opposite in sign.  If a +2.00D lens was required to neutralize the vertical meridian, then the power in the vertical meridian (90 degrees) is -2.00D. And if a +3.00 lens was required to neutralize the horizontal meridian then the power of the lens in the horizontal meridian is -3.00 (at 180 degrees).
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  STEPS IN NEUTRALIZATION 6.Convert to sphero-cylindrical formula: Convert your power cross into a spherocylindrical formula in the standard • in the example above (-2.00D at 90 and -3.00D at 180), the spherocylindrical formula is: -2.00D/-1.00x 90 • Conversion of Power Cross to Spherocylindrical formula: (the power of the more positive meridian) / - (difference between meridian) x (angle of more positive meridian)