# lens - glass the radius from the origin and the distance x...

This preview shows pages 1–2. Sign up to view the full content.

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 )?

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
This is the end of the preview. Sign up to access the rest of the document.

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 < 2, or n < √ 2....
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online