What is the expected power at axis 60° for the given RX of -1.50 -2.00 x090?

What is the expected power at axis 60° for the given RX of -1.50 -2.00 x090?

• A. -1.50
• B. -2.00
• C. 0.00
• D. -1.00

Answer: (B)

To determine the expected power at axis 60° for the prescription of -1.50 -2.00 x090, it’s essential to understand how the powers in a lens prescription interact with different axes. The prescription indicates that there is a spherical power of -1.50 D and a cylindrical power of -2.00 D oriented at 90 degrees. The cylindrical component affects how the lens refracts light based on the axis specified. At axis 90°, the full effect of the cylindrical power is utilized, while at any other axis, a component of the spherical and cylindrical powers needs to be calculated. At axis 60°, we need to find out how the cylindrical power of -2.00 D contributes in conjunction with the spherical power of -1.50 D. The angle difference between 90° (where the cylinder is fully effective) and 60° (the axis we are analyzing) is 30°. We can utilize the formula for the effective power at a given axis, which involves trigonometric functions to determine how much of the cylindrical power contributes at this new axis. The effective cylinder power at 60° can be calculated using the sine function since it establishes the projection of the cylindrical power. The effective cylindrical power

To determine the expected power at axis 60° for the prescription of -1.50 -2.00 x090, it’s essential to understand how the powers in a lens prescription interact with different axes.

The prescription indicates that there is a spherical power of -1.50 D and a cylindrical power of -2.00 D oriented at 90 degrees. The cylindrical component affects how the lens refracts light based on the axis specified. At axis 90°, the full effect of the cylindrical power is utilized, while at any other axis, a component of the spherical and cylindrical powers needs to be calculated.

At axis 60°, we need to find out how the cylindrical power of -2.00 D contributes in conjunction with the spherical power of -1.50 D. The angle difference between 90° (where the cylinder is fully effective) and 60° (the axis we are analyzing) is 30°. We can utilize the formula for the effective power at a given axis, which involves trigonometric functions to determine how much of the cylindrical power contributes at this new axis.

The effective cylinder power at 60° can be calculated using the sine function since it establishes the projection of the cylindrical power. The effective cylindrical power