A physicist is designing a contact lens. The material has is an index of refraction of n=1.4500. In order to yield the prescribed focal length, the physicist specifies the following dimensions: . . inner radius of curvature = +2.580 cm outer radius of curvature= +2.060 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)?
A physicist is designing a contact lens. The material has is an index of refraction of n=1.4500. In order to yield the prescribed focal length, the physicist specifies the following dimensions: . . inner radius of curvature = +2.580 cm outer radius of curvature= +2.060 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)?
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Transcribed Image Text:A physicist is designing a contact lens. The material has is an index of
refraction of n=1.4500. In order to yield the prescribed focal length, the
physicist specifies the following dimensions:
.inner radius of curvature = +2.580 cm
• outer radius of curvature = +2.060 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)?
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