## Abstract

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### References

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1. Plane Waves of Light: I. Electromagnetic Behavior, J.O.S.A. & R.S.I.,  15, p. 137; 1927.
[Crossref]
2. These sheets are of speculative character, since they depend upon the behavior of the free surface electricity under the influence of the electric field. It appears possible, however, that they may introduce very weak light of twice the incident frequency,which would be interesting, if true.
3. Equations are numbered as in the previous paper.
4. Physically this condition exists when polychromatic light is allowed to fall upon the surface—each color individually is reflected without regard to the presence of the others.
5. Except in one case. In that case, however, the Poynting vector has the same component as the velocity normal to the bounding surface: the two differ only in their tangential components. Hence our present argument is not affected.
6. The question is sometimes raised: If no net energy flows across the surface, how can there be any energy in the lower medium? The answer is, that the statement only applies to the steady-state condition. When light is first caused to shine on the surface, there is for a time a net flow into the metal; when the light is shut off, a net flow out of it. These, however, are transient effects. During the intermediate period, with equilibrium established, no net flow takes place.
7. I need hardly say that the ordinary laws of reflection and refraction are all contained in this table—not implicitly but explicitly. If they are not immediately obvious, it is only because the notation is unfamiliar.For example, the usual law of sines is contained in the equationa1=qq1r,ora1r=qq1:for when the media are dielectrics the velocities in the two media are v = p/q and v1 = p/q1, as was said in § 4. Moreover, in this case a1 is the sine of the angle of refraction, and r the sine of the angle of incidence. Hence a1/r = q/q1 becomes at once sin R/sin I = v1/v.Similarly the amplitude relations, when rewritten in trigonometric notation, are just the usual equations.The polarizing angle is that angle for which H2 = 0. From Table 2 it is easily seen to be Riven by κ=κ1 which easily reduces to cosI/cosR = v1/v, and this, together with the law of dines leads at once to the more familiar relation tan I = v1/v.Other common properties appear with equal ease.
8. The figures are drawn for a wave length 4346A, the optical constants being taken as N = 2.10, Ko = 3.80.
9. It need hardly be said that the intensity at a given depth is not independent of the angle of incidence; for that intensity depends upon how much light enters the conductor, as well as upon how rapidly it is absorbed. The former quantity depends markedly upon the angle of incidence; the latter only to a very small extent.
10. The same results can, of course, be obtained by summing the infinite series of internally reflected components.
11. “The Thickness of Spontaneously Deposited Photoelectrically Active Rubidium Films, Measured Optically,” J.O.S.A. & R.S.I.,  15, p. 374; 1927. I take this opportunity to ac: knowledge my indebtedness to Dr. Ives in connection with this study of Plane Waves of Light. Not only did it have its origin in the attempt to answer certain specific questions which he raised, but it has had throughout the encouragement of his continued interest and the benefit of his knowledge of experimental optics, concerning which I was (and by comparison still am) in a state of profound ignorance.
[Crossref]
12. If we wanted to find the angle of emergence in any particular case we could best do it by computing a1′ and c1′, and using their real parts only in the formulatanR′=−a1′/c1′.

#### 1927 (2)

Plane Waves of Light: I. Electromagnetic Behavior, J.O.S.A. & R.S.I.,  15, p. 137; 1927.
[Crossref]

“The Thickness of Spontaneously Deposited Photoelectrically Active Rubidium Films, Measured Optically,” J.O.S.A. & R.S.I.,  15, p. 374; 1927. I take this opportunity to ac: knowledge my indebtedness to Dr. Ives in connection with this study of Plane Waves of Light. Not only did it have its origin in the attempt to answer certain specific questions which he raised, but it has had throughout the encouragement of his continued interest and the benefit of his knowledge of experimental optics, concerning which I was (and by comparison still am) in a state of profound ignorance.
[Crossref]

#### J.O.S.A. & R.S.I. (2)

Plane Waves of Light: I. Electromagnetic Behavior, J.O.S.A. & R.S.I.,  15, p. 137; 1927.
[Crossref]

“The Thickness of Spontaneously Deposited Photoelectrically Active Rubidium Films, Measured Optically,” J.O.S.A. & R.S.I.,  15, p. 374; 1927. I take this opportunity to ac: knowledge my indebtedness to Dr. Ives in connection with this study of Plane Waves of Light. Not only did it have its origin in the attempt to answer certain specific questions which he raised, but it has had throughout the encouragement of his continued interest and the benefit of his knowledge of experimental optics, concerning which I was (and by comparison still am) in a state of profound ignorance.
[Crossref]

#### Other (10)

If we wanted to find the angle of emergence in any particular case we could best do it by computing a1′ and c1′, and using their real parts only in the formulatanR′=−a1′/c1′.

These sheets are of speculative character, since they depend upon the behavior of the free surface electricity under the influence of the electric field. It appears possible, however, that they may introduce very weak light of twice the incident frequency,which would be interesting, if true.

Equations are numbered as in the previous paper.

Physically this condition exists when polychromatic light is allowed to fall upon the surface—each color individually is reflected without regard to the presence of the others.

Except in one case. In that case, however, the Poynting vector has the same component as the velocity normal to the bounding surface: the two differ only in their tangential components. Hence our present argument is not affected.

The question is sometimes raised: If no net energy flows across the surface, how can there be any energy in the lower medium? The answer is, that the statement only applies to the steady-state condition. When light is first caused to shine on the surface, there is for a time a net flow into the metal; when the light is shut off, a net flow out of it. These, however, are transient effects. During the intermediate period, with equilibrium established, no net flow takes place.

I need hardly say that the ordinary laws of reflection and refraction are all contained in this table—not implicitly but explicitly. If they are not immediately obvious, it is only because the notation is unfamiliar.For example, the usual law of sines is contained in the equationa1=qq1r,ora1r=qq1:for when the media are dielectrics the velocities in the two media are v = p/q and v1 = p/q1, as was said in § 4. Moreover, in this case a1 is the sine of the angle of refraction, and r the sine of the angle of incidence. Hence a1/r = q/q1 becomes at once sin R/sin I = v1/v.Similarly the amplitude relations, when rewritten in trigonometric notation, are just the usual equations.The polarizing angle is that angle for which H2 = 0. From Table 2 it is easily seen to be Riven by κ=κ1 which easily reduces to cosI/cosR = v1/v, and this, together with the law of dines leads at once to the more familiar relation tan I = v1/v.Other common properties appear with equal ease.

The figures are drawn for a wave length 4346A, the optical constants being taken as N = 2.10, Ko = 3.80.

It need hardly be said that the intensity at a given depth is not independent of the angle of incidence; for that intensity depends upon how much light enters the conductor, as well as upon how rapidly it is absorbed. The former quantity depends markedly upon the angle of incidence; the latter only to a very small extent.

The same results can, of course, be obtained by summing the infinite series of internally reflected components.

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### Figures (12)

F. 9
F. 10
F. 11

Reflection at a glass-air boundary: electric vector perpendicular to the plane of incidence. The dotted lines in the upper part of the figure represent the damping vector.

F. 12

Reflection at a glass-air boundary: magnetic vector normal to the plane of incidence The dotted lines in the upper part of the figure represent the damping vector.

F. 13

Reflection at an air-platinum boundary: electric vector normal tothe plane of incidence. In the top part of the figure the vertical vectors, which do not change direction as the angle of incidence changes, represent the damping.

F. 14

Reflection at an air-platinum boundary: magnetic vector normal to the plane of incidence. In the top part of the figure the vertical vectors, which do not change direction as the angle of incidence changes, represent the damping.

F. 15
F. 16
F. 17
F. 18

Reflection at an air-rubidium boundary: electric vector normal to the plane of incidence. In the top part of the figure the vertical vectors, which do not change direction as the angle of incidence changes, represent the damping.

F. 19

Reflection at an air-rubidium surface: magnetic vector.normal to the plane of incidence In the top part of the figure the vertical vectors, which do not change direction as the angle of. incidence changes, represent the damping.

F. 20

### Tables (2)

Table 1 The equations of reflection and refraction. E normal to the plane of incidence

Table 2 The equations of reflection and refraction. H normal to the plane of incidence

### Equations (43)

$e i q ( α x + β y + γ z ) − ipt$
$H y + − H y − = − I x / c , H x + − H y − = + I y / c ,$
$E x = E l e i q ( a x + b y + c z ) − ipt , ⋯ ⋯ ⋯ ⋯ H z = H n ′ e i q ( a x + b y + c z ) − ipt .$
$E x 1 = E 1 l 1 e i q 1 ( a 1 x + b 1 y + c 1 z ) − i p 1 t , ⋯ ⋯ ⋯ ⋯ H z 1 = H 1 n 1 ′ e i q 1 ( a 1 x + b 1 y + c 1 z ) − i p 1 t .$
$E l e i q ( a x + b y ) − ipt = E 1 l 1 e i q 1 ( a 1 x + b 1 y ) − i p 1 t .$
$p = p 1 , q a = q 1 a 1 , q b = q 1 b 1 .$
$a 1 2 + b 1 2 + c 1 2 = 1 ,$
$q 1 a 1 = q a , q 1 b 1 = q b , a 2 = a , b 2 = b ;$
$a = r , a 1 = ( q / q 1 ) r , a 2 = r , b = 0 , b 1 = 0 , b 2 = 0 , c = − ( 1 − r 2 ) 1 / 2 , c 1 = ± ( 1 − q 2 r 2 / q 1 2 ) 1 / 2 , c 2 = + ( 1 − r 2 ) 1 / 2 .$
$q 1 a 1 = q a + i 0 , q 1 b 1 = 0 + i 0 , q 1 c 1 = R + i I ,$
$q 1 a 1 = q a + i 0 , q 1 b 1 = 0 + i 0 , q 1 c 1 = 0 + i I ,$
$l = 0 m = 1 n = 0 } ( E normal to plane of incidence ) , l ′ = 0 m ′ = 1 n ′ = 0 } ( H normal to plane of incidence ) ,$
$E = E 5 + E 7 , E 2 = E 6 + E 8 .$
$E 4 ′ / E 1 ′ = 2 k 1 / ( k 4 + k 1 ) , E 3 ′ / E 1 ′ = − ( k 4 − k 1 ) / ( k 4 + k 1 ) ; E 1 / E 5 = 2 k / ( k 1 + k ) , E 6 / E 5 = − ( k 1 − k ) / ( k 1 + k ) ; E 3 / E 7 = 2 k / ( − k 1 + k ) , E 8 / E 7 = ( k 1 + k ) / ( − k 1 + k ) .$
$E e i q ( a x + b y ) − ipt$
$E e − iqc ζ e i q ( a x + b y ) − ipt$
$E 1 ′ = E 1 e − i q 1 c 1 ζ , E 3 ′ = E 3 e i q 1 c 1 ζ ;$
$E 1 E = 2 k ( k 1 + k 4 ) X 2 ( k + k 1 ) ( k 1 + k 4 ) X 2 + ( k − k 1 ) ( k 1 − k 4 ) , E 2 E = ( k − k 1 ) ( k 1 + k 4 ) X 2 + ( k + k 1 ) ( k 1 − k 4 ) ( k + k 1 ) ( k 1 + k 4 ) X 2 + ( k − k 1 ) ( k 1 − k 4 ) , E 3 E = 2 k ( k 1 − k 4 ) ( k + k 1 ) ( k 1 + k 4 ) X 2 + ( k − k 1 ) ( k 1 − k 4 ) , E 4 ′ E = 4 k k 1 X ( k + k 1 ) ( k 1 + k 4 ) X 2 + ( k − k 1 ) ( k 1 − k 4 ) ,$
$X = e i q 1 c 1 ζ .$
$tan θ = component E normal to the plane of incidence component of E in the plane of incidence .$
$E ⊥ E ∥ = e i ϕ tan θ = ρ .$
$ρ = E ⊥ E ∥ = κ 1 − κ κ 1 + κ k 1 + k k 1 − k .$
$ρ = q c 1 − c q 1 q c 1 + c q 1 q 1 c 1 + q c q 1 c 1 − q c = q q 1 ( c 1 2 − c 2 ) + c c 1 ( q 1 2 − q 2 ) q q 1 ( c 1 2 − c 2 ) − c c 1 ( q 1 2 − q 2 ) .$
$c 1 2 − c 2 = r 2 ( q 1 2 − q 2 ) / q 1 2 ,$
$ρ = q q 1 r 2 + c c 1 q q 1 r 2 − c c 1 .$
$q 1 q = r c ( r + 1 ) [ ( ρ + 1 ) 2 − 4 r 2 ρ ] 1 / 2 ;$
$N + i K 0 = tan I 1 + sin I [ ( ρ + 1 ) 2 − 4 ρ sin 2 I ] 1 / 2 .$
$K = K 0 / N .$
$a = sin I , a 1 = q q 1 a , b = 0 , b 1 = 0 , c = − cos I , c 1 = − ( 1 − q 2 q 1 2 a 2 ) 1 / 2 .$
$a ′ = sin ( R − ϕ ) = a 1 cos ϕ + c 1 sin ϕ b ′ = 0 c ′ = − cos ( R − ϕ ) = − a 1 sin ϕ + c 1 cos ϕ ;$
$a 1 ′ = sin R ′ = q 1 q a ′ = a cos ϕ − ( q 1 2 q 2 − a 2 ) 1 / 2 sin ϕ . c 1 ′ = − cos R ′ = − ( 1 − a 1 ′ 2 ) 1 / 2 .$
$q 1 / q = a / sin 1 2 ϕ .$
$δ = ( I − R ) + ( I ′ − R ′ )$
$2 I = δ + ϕ .$
$N = sin 1 2 ( δ + ϕ ) / sin 1 2 ϕ .$
$N = sin 1 2 ( δ + ϕ ) sin 1 2 ϕ + ( K 0 2 − 1 ) sin 1 2 ( δ + ϕ ) sin 1 2 ϕ + N N 2 + K 0 2 sin 2 ( δ + ϕ ) + ⋯ ,$
$E 1 E = k − k 2 k 1 − k 2$
$E 2 E = − k 1 − k k 1 − k 2$
$H 1 H = κ − κ 2 κ 1 − κ 2$
$H 2 H = − κ 1 − κ κ 1 − κ 2$
$a1=qq1r,$
$a1r=qq1:$
$tanR′=−a1′/c1′.$