The Mirror Formula and Magnification

From similar triangles to one formula

Ray diagrams find an image by drawing. A formula finds it by arithmetic. Take a concave mirror, and an object AB standing on the axis with B on the axis. Two rays from A locate the tip A′ of the image A′B′. The ray parallel to the axis meets the mirror at M, at the height of A, and reflects through F. The ray to the pole reflects symmetrically.

Two pairs of triangles in Fig 5.1 are similar. The triangles A′B′F and MPF have equal angles at F and right angles at B′ and P. The triangles A′B′P and ABP have equal angles at P by the law of reflection, and right angles at B and B′. For paraxial rays, M is so close to the plane of the pole that PM equals AB. The rest is the sign convention.

The derivation used one drawing, but the formula holds for every case. Redraw the figure for a virtual image or a convex mirror, and the same steps give the same result. The signs take care of the positions. So one formula serves all mirrors. Put every known quantity in with its sign, and read the sign of the answer.

For a convex mirror \(f > 0\) and \(u < 0\), so \(1/v = 1/f - 1/u\) is a sum of two positive terms. It is larger than \(1/f\), so \(v\) is positive and smaller than \(f\). The image of any real object is therefore behind a convex mirror, between P and F, exactly as the ray diagrams showed.

How big, and which way up

The linear magnification \(m\) is the height of the image divided by the height of the object, both with their signs. The similar triangles A′B′P and ABP of Fig 5.1 give it at once. The image height is \(-A'B'\), since the image is below the axis, and the ratio of the triangles is \(B'P/BP = v/u\). So \(m = -v/u\).

The sign of \(m\) says which way up the image is. A negative \(m\) means an inverted image, and for a real object that image is real. A positive \(m\) means an erect image, and for a real object that image is virtual. The size of \(m\) says whether the image is larger, \(|m| > 1\), or smaller, \(|m| < 1\), than the object.

Walk the object inwards on the concave mirror, from \(70\,\text{cm}\). At \(70\,\text{cm}\) the image is small and inverted, at \(v = -28\,\text{cm}\), \(m = -0.40\). At \(40\,\text{cm}\), at C, it is the same size. At \(30\,\text{cm}\) it is twice the size at \(-60\,\text{cm}\), and at \(25\,\text{cm}\) four times the size at \(-100\,\text{cm}\). At F the image runs off to infinity. Just inside F it reappears behind the mirror, huge and erect: at \(15\,\text{cm}\), \(v = +60\,\text{cm}\) and \(m = +4\).

Distances from the focus, areas and lengths along the axis

Sometimes it is easier to measure from the focus. Let \(x\) be the distance of the object from F and \(y\) the distance of the image from F, both taken as positive lengths. Then the mirror formula becomes Newton's formula, \(xy = f^2\). A concave mirror with \(f = 20\,\text{cm}\) and an object \(5\,\text{cm}\) beyond F gives \(y = 400/5 = 80\,\text{cm}\). The image is \(20 + 80 = 100\,\text{cm}\) from the pole. The mirror formula agrees: \(u = -25\,\text{cm}\) gives \(v = -100\,\text{cm}\).

A small flat card facing the mirror has its width and its height each multiplied by \(m\). Its area is therefore multiplied by \(m^2\), the areal magnification. A square card of side \(1\,\text{cm}\), \(15\,\text{cm}\) in front of a concave mirror of focal length \(10\,\text{cm}\), has \(v = -30\,\text{cm}\) and \(m = -2\). Its image has an area of \(4 \times 1 = 4\,\text{cm}^2\).

A short rod lying along the axis is different. Its two ends are at slightly different \(u\), so their images are at slightly different \(v\). Differentiating the mirror formula gives \(dv/du = -v^2/u^2 = -m^2\), the longitudinal magnification. The minus sign says that the order of the ends is reversed. A rod \(1\,\text{mm}\) long at \(u = -30\,\text{cm}\) before a mirror with \(f = -20\,\text{cm}\) has \(m = -2\), so its image is \(4\,\text{mm}\) long.

The whole story in two graphs

The walk of the object in the ray-diagram sim can be drawn once as a graph. For a concave mirror with \(f = -20\,\text{cm}\), the formula gives \(v = -20u/(u + 20)\). The graph of \(v\) against \(u\) has two branches, divided by the dashed line \(u = f\). Beyond F the image is real, with \(v\) negative, and \(v\) runs to minus infinity as the object nears F. Inside F the image is virtual, with \(v\) positive, falling to zero at the pole.

Far from the mirror the real branch approaches the dashed line \(v = f\): an object at infinity images at F. The magnification tells the same story in the lower graph. It is \(-1\) at C, grows without limit near F, changes sign across F, and falls to \(+1\) at the pole. Very near the pole every spherical mirror behaves like a plane mirror, with \(v \approx -u\) and \(m \approx 1\).

A third graph is a straight line. The mirror formula reads \(1/v = -1/u + 1/f\). So \(1/v\) plotted against \(1/u\) is a line of slope \(-1\), cutting both axes at \(1/f\). If both intercepts are \(-0.05\,\text{cm}^{-1}\), then \(f = -20\,\text{cm}\), a concave mirror. The line meets \(1/v = 1/u\) at \(-0.025\,\text{cm}^{-1}\), which is \(u = v = -40\,\text{cm}\): the centre of curvature.

Now back to the dentist. The mirror is concave, with a radius of about \(4\,\text{cm}\), so \(f = -2\,\text{cm}\). The dentist holds it about \(1.5\,\text{cm}\) from the tooth, inside F. Then \(1/v = -1/2 + 1/1.5 = 1/6\), so \(v = +6\,\text{cm}\) and \(m = -6/(-1.5) = +4\). The tooth appears erect and four times larger. At arm's length your face is far beyond C, so the same mirror gives a small, inverted, real image.