lens - glass, the radius from the origin, and the distance...

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A half-cylinder of glass has radius R and index of refraction n . A ray of light lies in the plane of the figure as shown, and it is incident on the plane face at an angle θ .I tgoe s on to hit the curved surface; for what range of incident positions on the flat surface will the light be able to leave the cylinder when it first hits the curved surface? Let the position that it hits the flat surface be x as measured from the centerline of the flat. For what combination of n and θ will this range extend over the whole flat surface ( - R<x< + R )?
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Solution: Apply Snell’s law to see where the light will go inside the glass, then again Snell’s law with an exterior angle of 90 to find the angle at which total internal reflection starts to occur. sin θ = n sin φ and sin90 =1= n sin φ 0 Apply the law of sines to the triangle consisting of the path of the light ray inside the
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Unformatted text preview: glass, the radius from the origin, and the distance x . θ φ x R ' R sin( φ + π/ 2) = x sin φ This gives x = R sin φ sin( φ + π/ 2) = R 1 /n cos φ = R 1 /n q 1-sin 2 θ/n 2 = R p n 2-sin 2 θ . A similar construction on the other side (negative x ) gives R sin(-φ + π/ 2) =-x sin φ This gives-x = R sin φ sin(-φ + π/ 2) = R 1 /n cos φ = R 1 /n q 1-sin 2 θ/n 2 = R p n 2-sin 2 θ , so that the result is symmetric on the two sides of the centerline. This region can extend to the full distance where x = ± R if the denominator is one: p n 2-sin 2 θ = 1 , or sin 2 θ = n 2-1 . The sine can never be larger than one, so this can happen only if n 2 &lt; 2, or n &lt; √ 2....
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lens - glass, the radius from the origin, and the distance...

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