A plastic material with an index of refraction of 1.55 is used to form a lens. The radius of curvature of one lens surface is 1.10 m, and the radius of curvature of the other surface is 1.85 m. Use the lensmaker's equation to determine:

1. The magnitude $|f|$ of the focal length.
2. The power $|P|$ of the lens.

Using the lensmaker's equation with an index of refraction of 1.55, the focal length |f| of the lens is found to be approximately 4.94 meters, and the power |P| is approximately 0.20 diopters.

To determine the magnitude of the focal length |f| and the power |P| of a lens using a plastic material with an index of refraction of 1.55, the lensmaker's equation is utilized.

Given that the radius of curvature for one side of the lens is 1.10 m (which we will call R1), and for the other side is 1.85 m (R2), the equation takes the following form:

1/f = (n - 1) * (1/R1 - 1/R2)

Substituting the given values into the equation, we get:

1/f = (1.55 - 1) * (1/1.10 - 1/1.85)

1/f = 0.55 * (1/1.10 - 1/1.85)

= 0.55 * (0.909 - 0.541)

= 0.55 * 0.368

1/f = 0.2024

Therefore, the focal length |f| is:

f = 1 / 0.2024
≈ 4.94 m

To compute the lens power |P|, which is measured in diopters (D), we use the formula P = 1/f (where f is in meters).

So:

|P| = 1 / 4.94

≈ 0.20 D

This gives us the magnitude of the lens power.