EXAMPLE 1-4

Principal planes. Redo Section Exercises 1-3–1-6 for a complex lens system with principal planes separated by 1.0 cm = 0.01 m and net power P = +2.0 D.

Assume the object distances are referenced to the front principal plane. Then the image distances are the same as before, when referenced to the second principal plane, and are thus 1.0 cm greater than the answers to Section Exercises 1-3–1-6.

Section Exercises 1-11–1-14

Redo Section Exercises 1-3–1-6 for a lens system with net power P = +10 D.

1-11  u = −4.0 m, P = +10 D

ν = 1/9.75 D = 0.10256 m from the second principal plane

1-12  u = −1.0 m, P = + 10 D

ν = 1/9.00 D = 0.111 m from the second principal plane

1-13  u = −0.5 m, P = + 10 D

ν = 1/8.0 D = 0.125 m from the second principal plane

1-14  u = −0.4 m, P = + 10 D

ν = 1/7.50 D = 0.1333 m from the second principal plane

EXAMPLE 1-5

Principal planes and distant objects. The Gullstrand mathematical model closely approximates the human eye. (The Gullstrand model eye is discussed in Chapter 3.) The net power of the entire refracting system of the model eye is P = +58.64 D; the refractive index of the vitreous humor is 1.336, and the second (posterior) principal plane is located 1.602 mm behind the anterior corneal surface. Locate an image of a distant object as formed by the Gullstrand model eye with relaxed accommodation.

The image is located according to the vergence equation: 0 + 58.64 D = 1.336/ν, so ν = 0.02278 m, or 1.602 + 22.78 = 24.382 mm from the anterior corneal surface. For comparison, the axial length is 24.40 mm.

EXAMPLE 1-6

Effectivity of lenses. Suppose a lens of power P in air places the image of a distant object at a distance u to the right of the lens. To move the lens t meters to the right while preserving the image location, how must you adjust the power of the lens?

In its new position, the lens will form the image at a distance u2 = u − t to the right of the lens (as v = 0). Thus, the power P2 of the moved lens should be P2 = 1/(u − t) = 1/(1/P − t) = P/(1 − tP). Indeed, the same formula applies for divergent lenses, with the initial image location to the left of the lens, so long as the proper sign conventions are observed. This adjustment is referred to as the correction for the effectivity of lenses.

The effectivity of lenses comes into play, for example, when we convert a spectacle correction to the corresponding contact lens correction. For example, a hyperopic correction will place the image of a distant object at the far point of the eye (the location conjugate to the fovea with relaxed accommodation—see the Quick-Start Guide). In this case, the far point is behind the retina, and the effectivity calculation requires that the power of the correcting contact lens be greater than the power of the equivalent spectacles. For example, for a spectacle correction of +4.00 D worn at the standard vertex distance of 13.75 mm from the corneal apex, the power of the equivalent contact lens is +4.00 D/[1 − (0.01375 m)(4.00 D)] = +4.232 D, which suggests that the contact lens power should be increased by ¼ D. This correction is negligible for lenses weaker than about 3 D.