Light always reflects according to the law of reflection, regardless of whether the reflection occurs off a flat surface or a curved surface. Using reflection laws allows one to determine the image location for an object. The image location is the location where all reflected light appears to diverge from. Thus to determine this location demands that one merely needs to know how light reflects off a mirror. The image was merely that location where all reflected rays intersected. The use of the law of reflection to determine a reflected ray is not an easy task. For each incident ray, a normal line at the point of incidence on a curved surface must be drawn and then the law of reflection must be applied. A simpler method of determining a reflected ray is needed.

The simpler method relies on two rules of reflection for concave mirrors. They are:

Any incident ray traveling parallel to the principal axis on the way to the mirror will pass through the focal point upon reflection.

Any incident ray passing through the focal point on the way to the mirror will travel parallel to the principal axis upon reflection.

These two rules of reflection are illustrated in the diagram below.

The method for drawing ray diagrams for concave mirror is described below. The method is applied to the task of drawing a ray diagram for an object located *beyond* the center of curvature (C) of a concave mirror. Yet the same method works for drawing a ray diagram for any object location.

1. Pick a point on the top of the object and draw two incident rays traveling towards the mirror.

Using a straight edge, accurately draw one ray so that it passes exactly through the focal point on the way to the mirror. Draw the second ray such that it travels exactly parallel to the principal axis. Place arrowheads upon the rays to indicate their direction of travel.

2. Once these incident rays strike the mirror, reflect them according to the two rules of reflection for concave mirrors.

The ray that passes through the focal point on the way to the mirror will reflect and travel parallel to the principal axis. Use a straight edge to accurately draw its path. The ray that traveled parallel to the principal axis on the way to the mirror will reflect and travel through the focal point. Place arrowheads upon the rays to indicate their direction of travel. Extend the rays past their point of intersection.

3. Mark the image of the top of the object.

The image point of the top of the object is the point where the two reflected rays intersect. If your were to draw a third pair of incident and reflected rays, then the third reflected ray would also pass through this point. This is merely the point where all light from the top of the object would intersect upon reflecting off the mirror. Of course, the rest of the object has an image as well and it can be found by applying the same three steps to another chosen point.

4. Repeat the process for the bottom of the object.

The goal of a ray diagram is to determine the location, size, orientation, and type of image that is formed by the concave mirror. Typically, this requires determining where the image of the upper and lower extreme of the object is located and then tracing the entire image. After completing the first three steps, only the image location of the top extreme of the object has been found. Thus, the process must be repeated for the point on the bottom of the object. If the bottom of the object lies upon the principal axis (as it does in this example), then the image of this point will also lie upon the principal axis and be the same distance from the mirror as the image of the top of the object. At this point the entire image can be filled in.

Example Problem on solving a Concave Mirror Problem

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. Determine the image distance and the image size.

Like all problems in physics, begin by the identification of the known information.

_{o} = 4.0 cm | _{o} = 45.7 cm |

Next identify the unknown quantities that you wish to solve for.

_{i} = ??? | _{i} = ??? |

To determine the image distance, the mirror equation must be used. The following lines represent the solution to the image distance; substitutions and algebraic steps are shown.

_{i}

1/(15.2 cm) = 1/(45.7 cm) + 1/d_{i}

0.0658 cm^{-1} = 0.0219 cm^{-1 }+ 1/d_{i}

0.0439 cm^{-1} = 1/d_{i}

d_{i }= 22.8 cm |

The numerical values in the solution above were rounded when written down, yet un-rounded numbers were used in all calculations. The final answer is rounded to the third significant digit.

To determine the image height, the magnification equation is needed. Since three of the four quantities in the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown below.

_{i}/h

_{o}= - d

_{i}/d

_{o}

h_{i }/(4.0 cm) = - (22.8 cm)/(45.7 cm)

h_{i }= - (4.0 cm) • (22.8 cm)/(45.7 cm)

h_{i }= -1.99 cm |

Example Provided by Physics Classroom