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Physics · Ch 6 — Optics

Lateral Magnification in Spherical Mirrors

6.2.6

Lateral Magnification in Spherical Mirrors

Lateral (transverse) magnification is defined as the ratio of image height to object height, m=h′/hm=h'/h. Applying the Cartesian sign convention to the similar triangles used in deriving the mirror equation gives m=h′/h=−v/um=h'/h=-v/u; combining this with the mirror equation itself, mm can equivalently be written as m=ff−u=f−vfm=\dfrac{f}{f-u}=\dfrac{f-v}{f}. The sign of mm carries direct physical meaning: negative mm means a real, inverted image; positive mm means a virtual, erect image; and the magnitude of mm shows whether the image is enlarged (∣m∣>1|m|>1), the same size (∣m∣=1|m|=1), or diminished (∣m∣<1|m|<1). A distinct but related idea, longitudinal magnification (for an object extended along the axis rather than perpendicular to it), is the ratio of image length to object length, and is genera …

Misc Example 6.4Longitudinal magnification of a thin rod on the axis of a concave mirror

Worked out. A thin rod of length f/3 lies along the optical axis of a concave mirror of focal length f, positioned so its own real, elongated image just touches the rod at one end -- meaning that shared end must sit exactly at the mirror's centre of curvature, u' = R = 2f, so the near end of the rod sits at u = u' + f/3 = 2f + f/3 = 7f/3 (using the sign convention, the object distance actually used is -7f/3). Applying the mirror equation to locate the image of this near end, then defining the longitudinal magnification as the ratio of image length l' to object length l (here l = f/3), and solving the resulting equation for the magnification m shows m works out to 6/(m+3) in a self-consistent form that solves to m = 3/2. This illustrates that longitudinal magnification (along the axis) is generally different in value from the lateral (transverse) magnification of the same setup, since image distance …