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An engineer is designing a contact lens. The material has is an index of refraction of 1.55. In order to yield the prescribed focal length, the engineer specifies the following dimensions: inner radius of curvature = +2.42 cm outer radius of curvature = +1.98 cm where the inner radius of curvature describes the surface that touches the eye, and the outer radius of curvature describes the surface that first interacts with incoming light. What is the focal length of this contact lens (in cm)?

Respuesta :

Answer:20 cm

Explanation:

Given

Refractive index of material [tex]n=1.55[/tex]

Outer radius [tex]R_1=1.98\ cm[/tex]

Inner radius [tex]R_2=2.42\ cm[/tex]

using lens maker formula

[tex]\frac{1}{f}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})[/tex]

[tex]\frac{1}{f}=(1.55-1)(\frac{1}{1.98}-\frac{1}{2.42})[/tex]

[tex]f=\frac{10.895}{0.55}[/tex]

[tex]f=19.81\approx 20\ cm[/tex]

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