A technician wishes to produce the lens +2.00DS/+2.00DC x90 in toric form, in CR39 with a -5.00 base curve. After finishing the toroidal surface with the CR39 tool it is discovered that, in error, they have used a plastics material of refractive index 1.56. Which CR39 tool must now be used to work the front surface?
To determine which CR39 tool must be used to work the front surface, we need to consider the power and material properties.
Given information:
- Desired lens prescription: +2.00DS/+2.00DC x90 (toric form)
- Base curve for toroidal surface: -5.00 (from CR39 material)
Since the technician has mistakenly used a plastics material with a refractive index of 1.56 instead of CR39 (refractive index of 1.50), the incorrect material will affect the overall characteristics of the lens.
To compensate for the refractive index difference, we can make use of the lens thinning formula, which is:
D2 = D1 * (n1 / n2)
Where:
- D1 is the given lens power in the initial material (CR39)
- D2 is the desired lens power in the new material (plastics)
- n1 is the refractive index of the initial material (CR39)
- n2 is the refractive index of the new material (plastics)
In this case:
- D1 = +2.00DS (since the cylindrical power does not change when converting materials)
- n1 = 1.50 (refractive index of CR39)
- n2 = 1.56 (refractive index of plastics)
Plugging in the values, we have:
D2 = +2.00 * (1.50 / 1.56)
D2 ≈ +1.923DS
Therefore, the new desired lens power in the plastics material is approximately +1.923DS (ignoring the minus sign as cylindrical power does not change).
Now, to find the CR39 tool needed to work the front surface, we need to convert this new power back to the CR39 material.
Using the same lens thinning formula:
D2 = D1 * (n1 / n2)
- D1 is the desired lens power in the new material (plastics)
- D2 is the new lens power in the CR39 material
- n1 is the refractive index of the new material (plastics)
- n2 is the refractive index of the initial material (CR39)
In this case:
- D1 = +1.923DS
- n1 = 1.56 (refractive index of plastics)
- n2 = 1.50 (refractive index of CR39)
Plugging in the values, we have:
D2 = +1.923 * (1.56 / 1.50)
D2 ≈ +1.992DS
Therefore, the new lens power needed for the front surface in CR39 material is approximately +1.992DS (ignoring the minus sign as cylindrical power does not change).
To summarize, the technician needs to use the CR39 tool with a power of approximately +1.992DS to work the front surface after mistakenly using a plastics material with a refractive index of 1.56.