Page:EB1911 - Volume 23.djvu/16

Rh to $$\displaystyle\mathrm{MN}$$, then, by the law of reflection, angle $$\displaystyle\mathrm{ORT} = \mathrm{TRP}$$ or $$\displaystyle\mathrm{ORM} = \mathrm{PRN}$$. Hence draw $$\displaystyle\mathrm{OQ}$$ perpendicular to $$\displaystyle\mathrm{MN}$$, and produce it to $$\displaystyle\mathrm{S}$$, making $$\displaystyle\mathrm{QS} = \mathrm{OQ}$$; join $$\displaystyle\mathrm{SR}$$ and produce to $$\displaystyle\mathrm{P}$$. It is easily seen that $$\displaystyle\mathrm{PR}$$ and $$\displaystyle\mathrm{OR}$$ are equally inclined to $$\displaystyle\mathrm{RT}$$ (or $$\displaystyle\mathrm{MN}$$). A point-eye at $$\displaystyle\mathrm{P}$$ would see a point object $$\displaystyle\mathrm{O}$$ at $$\displaystyle\mathrm{S}$$, i.e. at a distance below the mirror equal to its height above. If the object be a solid, then the images of its corners are formed by taking points at the same distances below as the corners are above the mirror, and joining these points. The eye, however, sees the image perverted, i.e., in the same relation as the left hand to the right. Fig. 2 shows how an extended object is viewed in a mirror by a natural eye.

2. If $$\displaystyle\mathrm{A, B}$$ be two parallel plane mirrors and $$\displaystyle\mathrm{O}$$ a luminous point between them (fig. 3) to determine the images of $$\displaystyle\mathrm{O}$$ all the images must lie on the line (produced) $$\displaystyle\mathrm{PQ}$$ passing through $$\displaystyle\mathrm{O}$$ and perpendicular to the mirrors. Let $$\displaystyle\mathrm{OP} = p ,\ \mathrm{OQ} = q$$. Then if $$\displaystyle\mathrm{O'}$$ be the image of $$\displaystyle\mathrm{O}$$ in $$\displaystyle\mathrm{A} ,\ \mathrm{OO'} = 2p$$; now $$\displaystyle\mathrm{O'}$$ has an image $$\displaystyle\mathrm{O}$$ in $$\displaystyle\mathrm{B}$$, such that $$\displaystyle\mathrm{OO} = \mathrm{OQ} + \mathrm{QO} = q + q + 2p = 2p + 2q$$; similarly $$\displaystyle\mathrm{O}$$ has an image $$\displaystyle\mathrm{O}$$ in $$\displaystyle\mathrm{A}$$, such that $$\displaystyle\mathrm{OO} = 4p + 2q$$. In the same way $$\displaystyle\mathrm{O}$$ forms an image $$\displaystyle\mathrm{O_1}$$ in $$\displaystyle\mathrm{B}$$ such that $$\displaystyle\mathrm{OO_1} = 2q ;\ \mathrm{O_1}$$ has an image $$\displaystyle\mathrm{O_{11}}$$ in $$\displaystyle\mathrm{A}$$, such that $$\displaystyle\mathrm{OO_{11}} = 2p + 2q ;\ \displaystyle\mathrm{O_{11}}$$ has an image $$\displaystyle\mathrm{O_{111}}$$ in $$\displaystyle\mathrm{B}$$, such that $$\displaystyle\mathrm{OO_{111}} = 2p + 4q$$, and so on. Hence there are an infinite number of images at definite distances from the mirrors. This explains the vistas as seen, for example, between two parallel mirrors at the ends of a room.

3. If $$\displaystyle\mathrm{A, B}$$ be two plane mirrors inclined at an angle $$\displaystyle\theta$$, and intersecting at $$\displaystyle\mathrm{C}$$, and $$\displaystyle\mathrm{O}$$ a luminous point between them, determine the position and number of images.

Call arc $$\displaystyle\mathrm{OA} = \alpha ,\ \mathrm{OB} = \beta$$. The image of $$\displaystyle\mathrm{O}$$ in $$\displaystyle\mathrm{A}$$, i.e. $$\displaystyle a'$$, is such that $$\displaystyle\mathrm{O}a'$$ is perpendicular to $$\displaystyle\mathrm{CA}$$, and $$\displaystyle\mathrm{O}a' = 2\alpha$$. Also $$\displaystyle\mathrm{C}a' = \displaystyle\mathrm{CO}$$; and it is easily seen that all the images lie on a circle of centre $$\displaystyle\mathrm{C}$$ and radius $$\displaystyle\mathrm{CO}$$. The image $$\displaystyle a'$$ forms an image $$\displaystyle a$$ in $$\displaystyle\mathrm{B}$$ such that $$\displaystyle\mathrm{O}a = \mathrm{OB} + \mathrm{B}a'' = \beta + \mathrm{B}a' = \beta + \mathrm{O}B + \mathrm{O}a' = 2\beta + 2\alpha = 2\theta$$. Also $$\displaystyle a''$$ forms an image $$\displaystyle a$$ in $$\displaystyle\mathrm{A}$$ such that $$\displaystyle\mathrm{O}a = \mathrm{OA} + \mathrm{A}a' = 2\alpha + 2\theta$$. And generally $$\displaystyle\mathrm{O}a^{2n} = 2n\theta$$, $$\displaystyle\mathrm{O}a^{2n+1} = 2n\theta + 2\alpha$$. In the same way it can be shown that the image first formed in $$\displaystyle\mathrm{B}$$ gives foci of the general distances: $$\displaystyle\mathrm{O}b^{2n} = 2n\theta$$, $$\displaystyle\mathrm{O}b^{2n+1} = 2n\theta + 2\beta$$. The number of images is limited, for when any one falls on the arc $$\displaystyle ab$$ between the mirrors produced, it lies behind both mirrors, and hence no further image is possible. Suppose $$\displaystyle a^{2n}$$ be the first image to fall on this arc, then arc $$\displaystyle\mathrm{O}a^{2n} > \mathrm{OB}a$$, i.e. $$\displaystyle 2n\theta>\pi - \alpha$$ or $$\displaystyle 2n > (\pi-\alpha)/\theta .$$ Similarly if $$\displaystyle a^{2n + 1}$$ be the first to fall on $$\displaystyle ab$$, we obtain $$\displaystyle 2n+1 > \pi - \alpha/\theta$$. Hence in both cases the number of images is the integer next greater than $$\displaystyle\pi - \alpha/\theta$$. In the same way it can be shown that the number of images of the $$\displaystyle b$$ series is the integer next greater than $$\displaystyle \pi - \beta/\theta$$. If $$\displaystyle \pi/\theta$$ be an integer, then the number of images of each series is $$\displaystyle \pi/\theta$$, for $$\displaystyle \alpha/\theta$$ and $$\displaystyle \beta/\theta$$ are proper fractions. But an image of each series coincides; for if $$\displaystyle \pi/\theta = 2n$$, we have $$\displaystyle\mathrm{O}a^{2n} + \mathrm{O}b^{2n} = 2n\theta + 2n\theta = 2\pi$$ i.e. $$\displaystyle a^{2n}$$ and $$\displaystyle b^{2n}$$ coincide; and if $$\displaystyle \pi/\theta= 2n + 1$$, we have $$\displaystyle\mathrm{O}a^{2n + 1} + \mathrm{O}b^{2n + 1} = 4n\theta + 2\left(\alpha + \beta\right) = \left(4n + 2\right)\theta = 2\pi$$, i.e. $$\displaystyle a^{2n + 1}$$ and $$\displaystyle b^{2n+1}$$ coincide. Hence the number of images, including the luminous point, is $$\displaystyle 2\pi/\theta$$. This principle is utilized in the kaleidoscope (q.v.), which produces five images by means of its mirrors inclined at 60° (fig. 4). Fig. 5 shows the seven images formed by mirrors inclined at 45°.

4. To determine the reflection at a spherical surface. Let $$\displaystyle\mathrm{APB}$$ (fig. 6) be a section of a concave spherical mirror through its centre $$\displaystyle\mathrm{O}$$ and luminous point $$\displaystyle\mathrm{U}$$. If a ray, say $$\displaystyle\mathrm{UP}$$, meet the surface, it will be reflected along $$\displaystyle\mathrm{PV}$$, which is coplanar with $$\displaystyle\mathrm{UP}$$ and the normal $$\displaystyle\mathrm{PO}$$ at $$\displaystyle\mathrm{P}$$, and makes the angle $$\displaystyle\mathrm{VPO} = \mathrm{UPO}$$. Hence $$\displaystyle\mathrm{V0}/\mathrm{VP} = \mathrm{OU}/\mathrm{UP}$$. This expression may be simplified if we assume $$\displaystyle\mathrm{P}$$ to be very close to $$\displaystyle\mathrm{A}$$, i.e. that the ray $$\displaystyle\mathrm{UP}$$ is very slightly inclined to the axis. Writing $$\displaystyle\mathrm{A}$$ for $$\displaystyle\mathrm{P}$$, we have $$\displaystyle\mathrm{VO}/\mathrm{AV} = \mathrm{OU}/\mathrm{AU}$$; and calling $$\displaystyle\mathrm{AU} = u$$, $$\displaystyle\mathrm{AV} = v$$ and $$\displaystyle\mathrm{AO} = r$$, this reduces to $$\displaystyle u^{-1} + v^{-1} = r^{-1}$$. This formula connects the distances of the object and image formed by a spherical concave mirror with the radius of the mirror. Points satisfying this relation are called "conjugate foci," for obviously they are reciprocal, i.e. $$\displaystyle u$$ and $$\displaystyle v$$ can be interchanged in the formula. If $$\displaystyle u$$ be infinite, as, for example, if the luminous source be a star, then in $$\displaystyle v^{-1} = 2r^{-1}$$, i.e. $$\displaystyle v = \frac{1}{2}r$$. This value is called the focal length of the mirror, and the corresponding point, usually denoted by F, is called the “principal focus.” This formula requires modification for a convex mirror. If $$\displaystyle u$$ be always considered as positive ($$\displaystyle v$$ may be either positive or negative), $$\displaystyle r$$ must be regarded as positive with concave mirrors and negative with convex. Similarly the focal length, having the same sign as $$\displaystyle r$$, has different signs in the two cases.

In this formula all distances are measured from the mirror; but it is sometimes more convenient to measure from the principal focus. If the distances of the object and image from the principal focus be $$\displaystyle x$$ and $$\displaystyle y$$, then $$\displaystyle u = x + f$$ and $$\displaystyle v = y + f$$ (remembering that $$\displaystyle f$$ is positive for concave and negative for convex mirrors). Substituting these values in $$\displaystyle u^{-1} + v^{-1} = r^{-1}$$ and reducing we obtain $$\displaystyle xy = f^2$$. Since $$\displaystyle f^2$$ is always positive, $$\displaystyle x$$ and $$\displaystyle y$$ must have the same sign, i.e. the object and image must lie on the same side of the principal focus.

We now consider the production of the image of a small object placed symmetrically and perpendicular to the axis of a concave (fig. 7) and a convex mirror (fig. 8). Let $$\displaystyle\mathrm{PQ}$$ be the object and $$\displaystyle\mathrm{A}$$ the vertex of the mirror. Consider the point $$\displaystyle\mathrm{P}$$. Now a ray through $$\displaystyle\mathrm{P}$$ and parallel to the axis after meeting the mirror at $$\displaystyle\mathrm{M}$$ is reflected through the focus $$\displaystyle\mathrm{F}$$. The line $$\displaystyle\mathrm{MF}$$ must therefore contain the image of $$\displaystyle\mathrm{P}$$. Also a ray through $$\displaystyle\mathrm{P}$$ and also through the centre of curvature $$\displaystyle\mathrm{C}$$ of the mirror is reflected along the same path; this also contains the image of $$\displaystyle\mathrm{P}$$. Hence the image is at $$\displaystyle\mathrm{P}$$, the intersection of the lines $$\displaystyle\mathrm{MF}$$ and $$\displaystyle\mathrm{PC}$$. Similarly the image of any other point can be found, and the final image deduced. We notice that in fig. 6 the image is inverted and real, and in fig. 7 erect and virtual. The "magnification" or ratio of the size of the image to the object can be deduced from the figures by elementary geometry; it equals the ratio of the distances of the image and object from the mirror or from the centre of curvature of the mirror.

The positions and characters of the images for objects at varying