3.2 Magnification and the Lagrange Theorem |
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A ¼ f1 þ f2 df1f2
B¼ 1 df2
C¼ 1 df1
D¼ d:
3.2MAGNIFICATION AND THE LAGRANGE THEOREM
3.2.1 Transverse Magnification
Consider first a single refracting surface as in Figure 3.4. Let B, B0 be a pair of axial conjugate points, their distances from the surface being l and l0, respectively. We now place a small object at B and draw a paraxial ray from the top of the object to the vertex of the surface. The ray will be refracted there, the slope angles y and y0 being the angles of incidence and refraction. Hence ny ¼ n0y0, and therefore nh/l ¼ n0h0/l0. Multiplying both sides by y gives
hnu ¼ h0n0u0 |
(3-5) |
This important relationship is called the theorem of Lagrange, or sometimes the Smith–Helmholtz theorem. Because the h0, n0, and u0 on the right-hand side of one surface are, respectively, equal to the same quantities on the left-hand side of the next surface, it is clear that the product hnu is invariant for all the spaces between surfaces, including the object space and the image space. This product is called the Lagrange invariant or, more often nowadays, the optical invariant.
Since this theorem applies to the original object and also the final image, it is clear that the image magnification is given by
m ¼ h0=h ¼ nu=n0u0
For a lens in air, the magnification is merely u1=u0k (assuming that there are k surfaces in the system). The fact that the ratio of the nu values at the object
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Figure 3.4 The Lagrange relationship.
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Paraxial Rays and First-Order Optics |
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Figure 3.5 The Lagrange equation relationship for a distant object.
and at the image determines the magnification is one of the reasons it is usually preferred to trace a paraxial ray by the (y – nu) method.
In the case where the object is located at infinity, Eq. (3-5) is indeterminate
(u ¼ 0 and h ¼ 1). Figure 3.5 shows a very distant object where h0n0u0 ¼ ðh=lÞ nðluÞ. As l ! 1, the ratio ðh=lÞ tends toward tan up as ðluÞ becomes y, where y is the height on the first principal plane (see Section 3.3). Since y=u0 ¼ f 0 (posterior or rear focal length), then
h0 ¼ nn0 uy0 tan up ¼ nn0 f 0 tan up
where up is the slope of an entering parallel beam of light. Therefore the image height for an infinitely distant object is equal to the focal length times the angular subtense of the object.
DESIGNER NOTE
Recall that the anterior (front) focal length is f ¼ ðn=n0Þf 0, so using the preceding equation, we see that h0 ¼ f tan up. This can be interpreted by considering a ray passing through the anterior focal point with an angle up. It is evident that the ray will exit the lens parallel to the optical axis at a height h0. Since for small angles the tangent equals the angle in radians, we have two relationships to express focal length, namely,
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Both of these equations refer to paraxial rays. Although the following points will be covered in Chapter 4 and others, it should be observed that if the focal length varies with the lens aperture, the lens suffers coma. Should the focal length vary with obliquity, then distortion is present.
3.2 Magnification and the Lagrange Theorem |
65 |
3.2.2 Longitudinal Magnification
If an object has a small longitudinal dimension dl along the lens axis, or if it is moved along the axis through a small distance dl, then the corresponding longitudinal image dimension is dl0, and the longitudinal magnification m is given by m ¼ dl0/dl. By differentiating Eq. (3-4) we find
n0dl0=l02 ¼ n dl=l2
and multiplying both sides by y2 gives
n0dl0u02 ¼ n dl u2
This is the longitudinal equivalent of the Lagrange equation, and the product n dl u2 is also an invariant. The longitudinal magnification is found to be
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¼ dl0=dl ¼ mu2=n0u02 ¼ ðn0=nÞm2 |
(3-6) |
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so that for a lens in air, m ¼ m2. Hence longitudinal magnification is always positive, meaning that if the object is moved a short distance from left to right the image will move from left to right also. On the other hand, for a mirror, the signs of n and n0 are equal and opposite, so that when the object moves from left to right the image must move from right to left.
In situations where the ordinary magnification m is large, such as in a microscope objective, the longitudinal magnification will be very large, which explains the small depth of field noticed in a microscope. On the other hand, in a camera the ordinary magnification is small, so the longitudinal magnification is very small, accounting for the great depth of field noticed in most cameras.
Even when the object and image longitudinal dimensions are large, a useful expression for the longitudinal magnification can be derived. Using the Newtonian imaging equation, the magnifications of objects A and B, as shown in Figure 3.6, are related to their image distances z and z0 from the rear focal point of the lens by using this equation
z0A ¼ f 0mA and z0B ¼ f 0mB þ 81
Z ′ a
A B F2 A′ B′
Z ′ b
Figure 3.6 Longitudinal magnification.
66 |
Paraxial Rays and First-Order Optics |
When an object moves from A to B, the change in image distance A0B0 is then given by
A0B0 ¼ ðzA0 zB0 Þ ¼ f 0ðmA mBÞ:
The corresponding change in object distance AB is likewise given by
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Therefore, the large-scale longitudinal magnification A0B0=AB is given by
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When the longitudinal displacements AB and A0B0 are so small that magnification m hardly changes, then the above equation for longitudinal magnification reduces to m ¼ m2: The following example illustrates one application of m and m.
Example
Consider that a spherical object of radius ro is to be imaged as shown in Figure 3.7. The equation of the object is r2o ¼ y2o þ z2; where z is measured along the optical axis and is zero at the object’s center of curvature. Letting the surface sag as measured from the vertex plane of the object be denoted as
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The image of the object is expressed in the transverse or lateral direction by yi
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In a like manner, the image of the spherical object is expressed as ri (yi) / |
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Figure 3.7 Imaging of a spherical object.