Physics · Ch 6 — Optics

Focal Length of Lenses Out of Contact

6.6.7

Focal Length of Lenses Out of Contact

Two thin lenses of focal length $f_1,f_2$ separated by a distance $d$ cannot, in general, be replaced by a single equivalent thin lens for an object at a finite distance -- that requires the more involved theory of a 'thick lens', beyond this chapter's scope. But for the special case of an object at infinity (parallel incident rays only), tracing the small deviation angles $\delta_1=h_1/f_1$ and $\delta_2=h_2/f_2$ each lens produces on a parallel ray, and relating the ray's height at the second lens ( $h_2$ ) back to its height at the first ( $h_1$ ) through the geometry of straight-line travel across the separation $d$ , gives the combined focal length $\boxed{\dfrac{1}{F}=\dfrac{1}{f_1}+\dfrac{1}{f_2}-\dfrac{d}{f_1f_2}}$ , together with formulas locating the position of the single equivalent lens measured from each of the two real lenses (e.g. $PP_2=(d f_1)/f_1$ ... more precisely $PP_2=d f_1/(f_1)$ derived from the ratio $h_2/h_1$ , and similarly $PP_1=d f_2/f_2$ from the reverse ratio). This out-of-contact formula is valid * …

Figure 6.38Angle of deviation in lens

What this figure shows. A point object O on the axis sends an incident ray OA, striking a thin lens at a point A a small height h above the optical centre P; after refraction, this ray is deviated through an angle delta and crosses the axis again at the image point I. The small angles alpha (between the incident ray and the axis at O) and beta (between the refracted ray and the axis at I) satisfy delta = alpha + beta, which -- once alpha and beta are rewritten using PA=h, PO=-u and PI=v under the small-angle approximation -- simplifies neatly to the compact result delta = h/f, the starting point for a …

Figure 6.39Lens in out of contact

What this figure shows. Two thin lenses of focal length f1 and f2 sit a distance d apart on the same axis, both struck by a single incident ray parallel to the axis at height h1 on the first lens. The first lens deviates the ray by delta1 = h1/f1; the ray then travels the separation d, arriving at the second lens at a slightly different height h2, where it is deviated a second time by delta2 = h2/f2. Adding the two deviations (delta = delta1 + delta2) and eliminating h2 in favour of h1 and d using the triangle geometry of the ray's straight-line travel between the lenses is exactly how the combined out-of-contact focal-length formula 1/F = 1/f1 + 1/f2 - d/(f1 f2) is derived, valid stri …

Misc Example 6.18Final image position, nature and size for two separated lenses

Worked out. An object of height 5 mm sits 15 cm from a first convex lens of focal length 10 cm; a second lens of focal length 5 cm sits 40 cm beyond the first (55 cm from the object). Applying the lens equation to the first lens (u1 = -15 cm, f1 = 10 cm) gives v1 = 30 cm, so the first lens forms a real, inverted image 30 cm to its right, with magnification m1 = v1/u1 = 30/(-15) = -2, i.e. height h2 = 0.5 times (-2) = -1 cm (inverted, doubled in size). This image now sits 40 - 30 = 10 cm in front of the second lens, so u2 = -10 cm; applying the lens equation to the second lens (f2 = 5 cm) gives v2 = 10 cm, with magnification m2 = v2/u2 = 10/(-10) = -1. The final image height is h2' = h2 times m2 = (-1) times (-1) = +1 cm; since this final height is positive, the ultimate image is erect relative to the intermediate image (though inverted overall relative to the original obj …

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