To find the location of an image formed by a spherical mirror, we first use ray tracing, which is the technique of drawing rays and using the law of reflection to determine the reflected rays (later, for lenses, we use the law of refraction to determine refracted rays). The rules for ray tracing are summarized here for reference: We use ray tracing to illustrate how images are formed by mirrors and to obtain numerical information about optical properties of the mirror. From the geometry of the spherical mirror, note that the focal length is half the radius of curvature: We thus define the dimensionless magnification $$m$$ as follows: $\underbrace{m=\dfrac{h_i}{h_o}}_{\text{linear magnification}}. The Mirror Formula (also referred to as the mirror equation) gives us the relationship between the focal length (f), the distance of the object from the mirror (u) and the distance of the image from the mirror (v). \label{mirror equation}$, The mirror equation relates the image and object distances to the focal distance and is valid only in the small-angle approximation (Equation \ref{sma}). Check to see whether the answer makes sense. Figure $$\PageIndex{5}$$ shows a concave mirror and a convex mirror, each with an arrow-shaped object in front of it. Have questions or comments? Combined with some basic geometry, we can use ray tracing to find the focal point, the image location, and other information about how a mirror manipulates light. Write symbols and known values on the sketch. One of the solar technologies used today for generating electricity involves a device (called a parabolic trough or concentrating collector) that concentrates sunlight onto a blackened pipe that contains a fluid. Do the signs of object distance, image distance, and focal length correspond with what is expected from ray tracing? This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0). \begin{array}{rcl} \tanϕ=\dfrac{h_o}{d_o-R} \\ \tanϕ′=−\tanϕ=\dfrac{h_i}{R-d_i} \end{array}\right\} =\dfrac{h_o}{d_o-R}=−\dfrac{h_i}{R-d_i}\], $−\dfrac{h_o}{h_i}=\dfrac{d_o-R}{R-d_i}. A curved mirror, on the other hand, can form images that may be larger or smaller than the object and may form either in front of the mirror or behind it. Were we to move the object closer to or farther from the mirror, the characteristics of the image would change. It is valid only for paraxial rays, rays close to the optic axis, and does not apply to thick lenses. Parabolic mirrors focus all rays that are parallel to the optical axis at the focal point. An array of such pipes in the California desert can provide a thermal output of 250 MW on a sunny day, with fluids reaching temperatures as high as 400°C. Understanding the sign convention allows you to describe an image without constructing a ray diagram. [Image will be Uploaded Soon] On this page, we'll learn about the following: Type of spherical Mirrors Application: usage, Examples. To locate point $$Q′$$, drawing any two of these principle rays would suffice. Start with the equation for magnification (Equation \ref{mag}) and solving for $$d_i$$ and inserting the given values yields, \[d_i=−m d_o=−(0.032)(12\,cm)=−0.384\,cm \nonumber$, where we retained an extra significant figure because this is an intermediate step in the calculation. If the inside surface is the reflecting surface, it is called a concave mirror. Use the examples as guides for using the mirror equation. \label{eq61}\]. This heated fluid is pumped to a heat exchanger, where the thermal energy is transferred to another system that is used to generate steam and eventually generates electricity through a conventional steam cycle. The image distance $$d_i$$ is positive for real images and negative for virtual images. This mirror is a good approximation of a parabolic mirror, so rays that arrive parallel to the optical axis are reflected to a well-defined focal point. First make sure that image formation by a spherical mirror is involved. Example $$\PageIndex{2}$$: Image in a Convex Mirror. For a spherical mirror, the optical axis passes through the mirror’s center of curvature and the mirror’s vertex, as shown in Figure $$\PageIndex{1}$$. In deriving this equation, we found that the object and image heights are related by, $−\dfrac{h_o}{h_i}=\dfrac{d_o}{d_i}. a. A keratometer is a device used to measure the curvature of the cornea of the eye, particularly for fitting contact lenses. Taking the tangent of the angles $$θ$$ and $$θ′$$, and using the property that $$\tan(−θ)=−\tan θ$$, gives us, \[\left. Note that all incident rays that are parallel to the optical axis are reflected through the focal point—we only show one ray for simplicity. Although it was derived for a concave mirror, it also holds for convex mirrors (proving this is left as an exercise). For example, we show, as a later exercise, that an object placed between a concave mirror and its focal point leads to a virtual image that is upright and larger than the object. Note that the image distance here is negative, consistent with the fact that the image is behind the mirror. Example Problem #1 A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a concave mirror having a focal length of 15.2 cm. A ray that strikes the vertex of a spherical mirror is reflected symmetrically about the optical axis of the mirror (ray 4 in Figure $$\PageIndex{5}$$). Assume that all solar radiation incident on the reflector is absorbed by the pipe, and that the fluid is mineral oil. Consider a broad beam of parallel rays impinging on a spherical mirror, as shown in Figure $$\PageIndex{8}$$. Are the object and image distances reasonable. A ray traveling along a line that goes through the center of curvature of a spherical mirror is reflected back along the same line (ray 3 in Figure $$\PageIndex{5}$$). In fact, we already used ray tracing above to locate the focal point of spherical mirrors, or the image distance of flat mirrors. This means the focal point is at infinity, so the mirror equation simplifies to. What does it mean to have a negative radius of curvature? The highest point of the object is above the optical axis, so the object height is positive. Step 6. What is the amount of sunlight concentrated onto the pipe, per meter of pipe length, assuming the insolation (incident solar radiation) is 900 W/m. When this approximation is violated, then the image created by a spherical mirror becomes distorted. We choose to draw our ray from the tip of the object. How does the focal length of a mirror relate to the mirror’s radius of curvature? The radius r for a concave mirror is a negative quantity (going left from the surface), and this gives a positive focal length, implying convergence. It assigns positive or negative values for the quantities that characterize an optical system. To keep track of the signs of the various quantities in the mirror equation, we now introduce a sign convention. Notice that we have been very careful with the signs in deriving the mirror equation. Thus, the focal point is virtual because no real rays actually pass through it; they only appear to originate from it. Step 1. Symmetry is one of the major hallmarks of many optical devices, including mirrors and lenses. The difficulty is that, because these rays are collinear, we cannot determine a unique point where they intersect. The radius of curvature found here is reasonable for a cornea. The pole serves as the origin. Thus, triangle $$CXF$$ is an isosceles triangle with $$CF=FX$$. \label{eq55}$, Combining Equation \ref{eq51} and \ref{eq55} gives, $\dfrac{d_o}{d_i}=\dfrac{d_o−R}{R−d_i}.$, \dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{2}{R}. However, as discussed above, in the small-angle approximation, the focal length of a spherical mirror is one-half the radius of curvature of the mirror, or $$f=R/2$$. The mass $$m$$ of the mineral oil in the one-meter section of pipe is, \[ \begin{align*} m &= ρV = ρπ\left(\dfrac{d}{2}\right)^2(1.00\,m) \nonumber \\[4pt] &=(8.00×10^2kg/m^3)(3.14)(0.0100\,m)^2(1.00\,m) \nonumber \\[4pt] &=0.251\,kg \end{align*}, Therefore, the increase in temperature in one minute is, \begin{align*} \Delta T&= \dfrac{Q}{mc} \nonumber \\[4pt] &=\dfrac{(1130\,W)(60.0\,s)}{(0.251\,kg)(1670\,J⋅kg/°C)} \nonumber \\[4pt] &=162°\end{align*}. We can extend the mirror equation to the case of a plane mirror by noting that a plane mirror has an infinite radius of curvature.