Abstract

The maximum is found for the deviation of light passing through a transparent wedge of refractive index n and wedge angle α. The methods are conceptual and geometric, and they require very little calculation. There turn out to be two qualitatively different ray path candidates for maximum deviation, and the geometric approach leads naturally to a criterion involving n and α that decides between the two candidates. Finding the maximum deviation is equivalent to finding the outer radius of a circular halo.

© 2008 Optical Society of America

Full Article  |  PDF Article

References

  • View by:
  • |
  • |
  • |

  1. H. S. Uhler, “On the deviation produced by prisms,” American Journal of Science, Fourth Series 35, 389-423 (1913).
  2. W. Tape, “Analytic foundations of halo theory,” J. Opt. Soc. Am. 70, 1175-1192 (1980).
    [CrossRef]
  3. W. Tape and J. Moilanen, Atmospheric Halos and the Search for Angle x (American Geophysical Union, 2006).
  4. M. Born and E. Wolf, Principles of Optics (Pergamon Press, 1975).

1980 (1)

1913 (1)

H. S. Uhler, “On the deviation produced by prisms,” American Journal of Science, Fourth Series 35, 389-423 (1913).

Born, M.

M. Born and E. Wolf, Principles of Optics (Pergamon Press, 1975).

Moilanen, J.

W. Tape and J. Moilanen, Atmospheric Halos and the Search for Angle x (American Geophysical Union, 2006).

Tape, W.

W. Tape, “Analytic foundations of halo theory,” J. Opt. Soc. Am. 70, 1175-1192 (1980).
[CrossRef]

W. Tape and J. Moilanen, Atmospheric Halos and the Search for Angle x (American Geophysical Union, 2006).

Uhler, H. S.

H. S. Uhler, “On the deviation produced by prisms,” American Journal of Science, Fourth Series 35, 389-423 (1913).

Wolf, E.

M. Born and E. Wolf, Principles of Optics (Pergamon Press, 1975).

American Journal of Science, Fourth Series (1)

H. S. Uhler, “On the deviation produced by prisms,” American Journal of Science, Fourth Series 35, 389-423 (1913).

J. Opt. Soc. Am. (1)

Other (2)

W. Tape and J. Moilanen, Atmospheric Halos and the Search for Angle x (American Geophysical Union, 2006).

M. Born and E. Wolf, Principles of Optics (Pergamon Press, 1975).

Cited By

OSA participates in CrossRef's Cited-By Linking service. Citing articles from OSA journals and other participating publishers are listed here.

Alert me when this article is cited.


Figures (13)

Fig. 1
Fig. 1

(a) Simulation of the common 22 ° halo. Theoretically the halo occupies the annular region between the concentric circles Δ = Δ min and Δ = Δ max , but the halo is exceedingly weak near the outer boundary Δ = Δ max . (b) Simulation of the upper sunvex, upper suncave, and lower sunvex 22 ° Parry arcs for a sun elevation of 16 ° . Since the ray paths for these arcs are the same as ray paths for the 22 ° halo, these arcs must be subsets of the annular region that is the 22 ° halo.

Fig. 2
Fig. 2

Light ray path through a transparent wedge. The deviation Δ is the angle between the entry ray and exit ray. The angle α is the wedge angle.

Fig. 3
Fig. 3

Ray path through a wedge of refractive index n at left, and corresponding “vee” P A Q at right. The points P, A, and Q are the light points of the entry, internal, and exit rays to the wedge; the rays themselves are in the directions of the vectors P O , A O , and Q O . Snell’s law is expressed by the fact that the segments A P and A Q at the right are in directions normal to the entry and exit faces of the wedge, and that they do not penetrate the unit sphere S. The points P, A, and Q, and hence the segments A P and A Q , are in the plane of the paper, but the origin O, and hence the ray path, need not be.

Fig. 4
Fig. 4

(a) The points (white) of the unit sphere S that are visible from A. One such point is Q; its distance from A is less than or equal to the length m of a tangent to the sphere drawn from A. Also shown is a line segment through A normal to S. (b) Typical vee P A Q . The points P and Q are visible from A. For a given refractive index n and wedge angle α, the maximum deviation problem is equivalent to finding the maximum base length | PQ | for vees with | A | = n and PAQ = α . (c)–(f) Some special vees, all with n = 1.31 (ice) and α = 40 ° .

Fig. 5
Fig. 5

Rabbit ears TV antenna serving as a vee. In this context the maximum deviation problem amounts to positioning the feet of the vee as far apart as possible on the beach ball, while keeping the apex and apex angle fixed. The legs of the vee extend or contract as needed.

Fig. 6
Fig. 6

(a) Typical vee with apex A and apex angle α and lying in the prescribed plane N. (b) Same but showing the section in the plane N. The angle D ( t ) is the projected deviation. (c) The (signed) angle t and the arc length s ( t ) . (d) The vee when t = α / 2 ; it is isosceles. (e) The vee when t = t 0 ; it is tangential. (f) Illustrating that the speed s ( t ) increases with | t | . The spokes emanating from A have equally spaced t-values, whereas the spacing of the corresponding points on the circle N S increases with | t | . As a result, the minimum and maximum of D ( t ) occur at the isosceles and tangential vees, respectively.

Fig. 7
Fig. 7

Three tangential vees, all with the same apex A, the same tangential foot P, and the same apex angle.

Fig. 8
Fig. 8

The same three vees of Fig. 7, but detached from the sphere. The central tangential vee P A C has the shortest possible second leg, and the doubly tangential vee P A T has the longest. In (c), where the three vees have been superimposed on each other, the second foot Q of the generic tangential vee therefore lies on the segment C T .

Fig. 9
Fig. 9

Geometric solution to the maximum deviation problem. The segment A P of length m = n 2 1 is tangent to the unit circle C at P. The point T is the result of rotating P through angle α about A, and C is the intersection of the segment A T with C. The larger of d C and d T is the maximum base length, which gives Δ max by Eq. (1).

Fig. 10
Fig. 10

Dependence of the maximum deviation problem on α. (a)  0 < α < α 2 . The point T is within the unit circle C. (b)  α = α 2 . The point T is on C. (c)  α 2 < α < π . The point T is outside C, and the maximum deviation problem is empty. ( m = 0.85 )

Fig. 11
Fig. 11

Calculating Δ C . See Prop. 2.

Fig. 12
Fig. 12

Deciding between C and T. The points of the line A through A and T that are closer to P than is T are those between U and T. When U is inside the unit circle C, as at the right, the maximum base length occurs at C, since C is then farther from P than is T. When U is outside C (but T is not), the maximum occurs at T.

Fig. 13
Fig. 13

Dependence of the maximum deviation problem on α; max at C versus max at T. (a)  0 < α < α 1 . The point U is inside C and the maximum is at C. (b)  α = α 1 . The point U is on C and the maximum is at both C and T. (c)  α 1 < α α 2 . The point U is outside C and the maximum is at T. ( m = 1.18 )

Equations (13)

Equations on this page are rendered with MathJax. Learn more.

d 2 = sin Δ 2 .
r D ( t ) = s ( t ) s ( t α ) ,
r D ( t ) = s ( t ) s ( t α ) { > 0 if     α t < t 0 = s ( t ) s ( t + α ) > 0 if     α / 2 < t α .
tan α 2 2 = 1 m .
m sin α 2 = cos α 2
cos ( α + Δ C ) = cos α m sin α ,
0 α + Δ C π .
sin Δ T 2 = m sin α 2 .
d T d C if     0 α α 1 ,
d C d T if     α 1 α α 2 ,
m sin α 2 = cos 3 α 2
Δ max = Δ C = α + cos 1 ( cos α m sin α ) .
Δ max = Δ T = 2 sin 1 ( m sin α 2 ) .

Metrics