Abstract

Lloyd’s mirror experiment is applied to testing flatness of large surfaces. Because of the grazing incidence, even rough surfaces provide the characteristic interference pattern. In the case of a perfectly flat surface, that pattern consists of narrowly and equally spaced two beam interference fringes. Departures from flatness are reflected in changes of the fringe spacing. Moiré techniques are used for visualizing and measuring these changes in the deviation from straightness of the moiré fringes. Qualitative examples are given.

© 1967 Optical Society of America

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References

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  1. R. N. Wolfe, J. Opt. Soc. Am. 38, 706 (1948);J. Opt. Soc. Am. 40, 143 (1950).
    [CrossRef]
  2. G. A. Sciammarella, Exp. Mech. 5, 154 (1965).
    [CrossRef]
  3. J. Ebbeni, Bull. Acad. Roy. Belgium 50, 706 (1964).

1965 (1)

G. A. Sciammarella, Exp. Mech. 5, 154 (1965).
[CrossRef]

1964 (1)

J. Ebbeni, Bull. Acad. Roy. Belgium 50, 706 (1964).

1948 (1)

Ebbeni, J.

J. Ebbeni, Bull. Acad. Roy. Belgium 50, 706 (1964).

Sciammarella, G. A.

G. A. Sciammarella, Exp. Mech. 5, 154 (1965).
[CrossRef]

Wolfe, R. N.

Bull. Acad. Roy. Belgium (1)

J. Ebbeni, Bull. Acad. Roy. Belgium 50, 706 (1964).

Exp. Mech. (1)

G. A. Sciammarella, Exp. Mech. 5, 154 (1965).
[CrossRef]

J. Opt. Soc. Am. (1)

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Figures (7)

Fig. 1
Fig. 1

Geometry of Lloyd flatness interferometer (z axis perpendicular to drawing plane).

Fig. 2
Fig. 2

Lloyd’s mirror interferometer to measure flatness.

Fig. 3
Fig. 3

Relation between coordinates on Lloyd’s interferogram and on test surface.

Fig. 4
Fig. 4

Relation between Lloyd moiré fringes and surface figure.

Fig. 5
Fig. 5

Lloyd’s interferometer with glass strip loaded from below to produce different curvatures.

Fig. 6
Fig. 6

Moiré interference fringes from Lloyd interferogram of glass strip (see Fig. 5) loaded with three different weights. The moiré fringes show effect of different sags. H = total height of illuminated field, h = height of moiré grating (see Technique A), t = height corresponding to direct reflection from surface, s = diffraction fringes from edge. (a) load: 0 g, sag = 0, (b) load 85 g, sag = 0.016 cm, (c) load 184 g, sag = 0.038 cm, (d) load 368 g, sag = 0.080 cm.

Fig. 7
Fig. 7

Comparison of Fizeau interferogram and Lloyd moiré interferogram (Technique B). (a) and (b) 30-cm mirror, 8.5 λ convex. (c) and (d) The same mirror after final polishing, 0.5 λ convex.

Tables (1)

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Table I Applied Loads and Test Pressures

Equations (40)

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( Q 1 M ) 2 = a 0 2 + ( y n - d 0 / 2 ) 2 , ( Q 2 M ) 2 = a 0 2 + ( y n + d 0 / 2 ) 2 .
( Q 2 M ) 2 - ( Q 1 M ) 2 = 2 y n d 0 = ( Q 2 M - Q 1 M ) ( Q 2 M + Q 1 M ) .
( Q 2 M + Q 1 M ) = [ a 0 2 + ( y n - d 0 / 2 ) 2 ] 1 2 + [ a 0 2 + ( y n + d 0 / 2 ) 2 ] 1 2
( Q 2 M + Q 1 M ) = 2 a 0 [ 1 + y n 2 / a 0 2 ] 1 2 .
( Q 2 M + Q 1 M ) = 2 a 0 [ 1 + y n 2 / 2 a 0 2 ] .
a 0 70 y n , max ,
y n × d 0 = ( Q 2 M - Q 1 M ) × a 0 [ 1 + 1 × 10 - 4 × y n 2 / y n , max 2 ] .
P n = a 0 λ / d 0 = λ / 2 .
Δ y = M n - M α .
P = P x ( α , h ) = [ λ a 0 / d x ( α , h ) ] .
2 [ d 0 / 2 + b ] cos α = 2 [ d 0 / 2 + x 1 * tan α ] cos α .
1 2 d x = 1 2 d 0 + x 1 * α + h ,
h = [ 4 ( x 0 - x 1 * ) 2 / 8 R ] .
h = [ ( x 0 - x 1 ) 2 / 2 R ] .
α = 2 h / ( x 0 - x 1 ) .
1 2 ( d x - d 0 ) = h [ 2 x 1 / ( x 0 - x 1 ) + 1 ] = h [ ( x 0 + x 1 ) / ( x 0 - x 1 ) ] .
( d x - d 0 ) = 6 h ,
h = 1 2 d 0 [ ( 1 / k ) - 1 ] [ ( x 0 - x 1 / x 0 + x 1 ) ] ,
h max = d 0 [ ( 1 / k ) - 1 ] .
S = Δ P [ d ( h ) ] Δ h .
Δ P Δ h = P d x × d x h ,
Δ P Δ h = ( λ a 0 / d x 2 ) × 6.
S = ( P 0 / d x ) × 6.
S = ( P 0 × 6 ) / ( d 0 ± 6 h ) .
y n + ( d 0 / 2 ) a 0 = ( d 0 / 2 ) x , x = a 0 x ( d 0 / 2 ) y n + ( d 0 / 2 ) , x = a 0 1 [ ( 2 y n / d 0 ) + 1 ] .
x = x - ( l / 2 ) - f , x = a 0 [ ( 2 y n / d 0 ) + 1 ] - ( l / 2 ) + f ) .
k = P x / b .
sin φ = sin β ( 1 + k 2 - 2 k cos β ) 1 2 ,
k 1 , 2 = cos β ± sin β cotan φ .
Δ M M = Δ P P x = P x - P 0 P x = [ 1 - ( b / P x ) ]
[ ( 1 / k ) - 1 ] = - ( Δ M / M ) .
M = b 2 sin ( β / 2 ) ;     P 0 = b ,
( 1 / M 2 ) = ( 1 / b 2 ) + ( 1 / P 0 2 ) - ( 1 / 2 b P 0 ) cos β ;     P 0 b .
k 1 = 1 + β cot φ ;     φ < 90 ,
k 2 = 1 - β cot φ ;     φ > 90 ,
[ ( 1 / k ) - 1 ] = + cot φ 1 ± β cot φ .
k 1 = cos 10 ° + ( sin 10 ° cot 81 ° ) = [ 0.985 + ( 0.174 × 0.158 ) ] = 1.0124.
P x 1 = b × 1.0124 = 2.0248 × 10 - 2 cm , d 0 1 = a 0 × λ / P x 1 = 80.6 × 10 - 5 / 2.025 × 10 - 2 = 0.23 cm .
( 1 / P 0 2 ) - ( 2 × 50 ) / ( P 0 ) - 50 2 - 0.5 2 = 0 ,
1 / P 0 = + 50 ± ( 50 2 + 0.5 2 - 50 2 ) 1 2 = 50 ± 0.25.

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