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A lens is formed from a plastic material that has an index of refraction of 1.55. If the radius of curvature of one surface is 1.05 m and the radius of curvature of the other surface is 1.85 m, use the lensmaker's equation to calculate:

1. The magnitude of the focal length [tex]||[/tex].
2. The power [tex]|P|[/tex] of the lens.

Answer :

The magnitude of the focal length is approximately 4.41 m, and the power of the lens is approximately 0.2265 D.

The lensmaker's equation relates the focal length (f), the radius of curvature (R), and the refractive index (n) of a lens. The equation is given by:
1/f = (n - 1) * ((1/R1) - (1/R2))
Where R1 is the radius of curvature of the first surface of the lens, R2 is the radius of curvature of the second surface of the lens, and n is the refractive index of the lens material.
In this case, the refractive index (n) of the lens material is given as 1.55. The radius of curvature of one surface (R1) is 1.05 m, and the radius of curvature of the other surface (R2) is 1.85 m.
To find the focal length, we can substitute the given values into the lensmaker's equation:
1/f = (1.55 - 1) * ((1/1.05) - (1/1.85))
Simplifying the equation:
1/f = 0.55 * (0.9524 - 0.5405)
1/f = 0.55 * 0.4119
1/f ≈ 0.2265
To calculate the magnitude of the focal length (|f|), we take the reciprocal of both sides of the equation:
|f| ≈ 1/0.2265
|f| ≈ 4.41 m
Therefore, the magnitude of the focal length is approximately 4.41 m.
The power (P) of a lens is given by the equation:
P = 1/f
Substituting the value of the focal length into the equation:
P = 1/4.41
P ≈ 0.2265
Therefore, the power of the lens is approximately 0.2265 diopters (D).
In summary, the magnitude of the focal length is approximately 4.41 m, and the power of the lens is approximately 0.2265 D.

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