The purpose of most optical instruments is to produce an image.  Images can be formed by reflection and refraction.

What is an image?

We see objects, because they either emit or reflect light.  We visually identify objects by the pattern of light that they emit or reflect.  We gather some portion of that light on a detector (our eyes).  In interpreting the pattern, we implicitly assume that the light traveled in a straight line from the object to the detector.

 A real image of an object produces the same pattern of light as the object does somewhere in space.  Some portion of the light from the real image reaches our detector along a straight-line path.  The detector cannot distinguish between light coming from the object and light coming from the image.  We interpret the same patterns in the same way. A virtual image is the apparent position from which a pattern of light reaches our detector, if we make the assumption that it has traveled from its source to the detector along a straight-line path.  Virtual images are formed when light from an object or from an image is reflected or bend on the way to the detector.
Mirrors can produce real and virtual images by reflection.  A flat mirror only produces virtual images.

Light rays from the object reflect off the mirror before hitting a detector.  For the detector, the apparent position of the object is behind the mirror.  Let xo denote the perpendicular distance of the object from the mirror surface, and let xi denote the perpendicular distance of the image from this surface.  If we let distances in front of the mirror be positive and distances behind the mirror be negative, then we have xo + xi = 0.  For mirrors, negative image distances are associated with virtual images and positive image distances are associated with real images.

For each ray θi = θr.

Multiple images can be formed by combinations of flat mirrors.  In the picture to the right, light bouncing off one or both of the mirrors can reach the eye along three different paths.  The detector sees three virtual images.
Problem:

Determine the minimum height of a vertical, flat mirror in which a person 5'10'' in height can see his or her full image.

 Solution: The height of the mirror must be half the height of the person, in this case 2'11''.

Problem:

Two flat mirrors have their reflecting surfaces facing one another, with the edge of one mirror in contact with he edge of the other, so that the angle between the mirrors is α.  When an object is placed between the mirrors, a number of images are formed.  In general, if the angle α is such that nα = 360o, where n is an integer, the number of images formed is n - 1.  Graphically find all the image positions for the case n = 6, when a point object is between the mirrors, but not on the angle bisector.

 Solution: I1 and I2 are the images of O formed by mirror 1 and mirror 2, respectively.  I3 is an image of I2 formed by mirror 1 and I4 is an image of I1 formed by mirror 2.  We can view I5 as the image of I4 formed by mirror 1 or the image of I3 formed by mirror 2.

#### Curved Mirrors

Light rays from a distant star are nearly perfectly parallel to each other when they reach earth.  To produce a bright image of the star we want to gather as much light from the star as possible and bring it together in one spot, called a focus.  A mirror with a parabolic surface can perfectly focus incoming light rays parallel to its axis.  It produces a real image in the focal point of the parabola.  A parabola is the locus of all points, which are equidistant from a line and a point (the focal point).  All light rays from a distant source approaching the parabola parallel to its symmetry axis and reflecting from its surface have the same minimum travel time to the focal point of the parabola.  By Fermat's principle light will therefore take all these paths, and the rays will come together at the focus.

A concave mirror with a spherical surface focuses light similarly to a parabolic mirror as long as the angle subtended by the spherical mirror section is small.  A spherical mirror has a radius of curvature R and a focal length f = R/2.  The figure on the right compares a parabolic and a concave spherical surface with the same focal length f.  Near the optical axis the curves are nearly identical.

If an object is placed in front of a spherical mirror at an object distance xo, then an image is formed at an image distance xi, where xo and xi satisfy the mirror equation,

1/xo + 1/xi = 1/f.

Sign conventions for the mirror equation:

xi is positive for a real image in front of the mirror surface, and xi is negative for a virtual image behind the mirror surface.  xo and xi are the perpendicular distances from the center plane of the mirror as shown in the drawing on the right.

The focal length f and the radius of curvature R = 2f are positive for a concave mirror and negative for a convex mirror.  For a concave mirror the reflecting surface bulges inward, and for a convex mirror the reflecting surface bulges outward.

We can determine the positions and sizes of images of points formed by spherical mirrors geometrically by drawing ray diagrams.  Only two rays must be drawn.

Rules for drawing ray diagrams:

 Draw the spherical mirror surface of radius R.  Draw a line through R, which crosses the spherical surface near its center.  This is the optical axis.  Mark R and f along the optical axis. Draw the object in front of the mirror surface.  Draw an incident ray parallel to the optical axis from a point on the object to the mirror, and a reflected ray from the mirror through f.  (The reflected ray, or an extension of the reflected ray must pass through f.) Draw a second incident ray through f, and a reflected ray parallel to the optical axis.  (The incident ray, or an extension of the incident ray must pass through f.) The intersection of the two reflected rays, or, for divergent rays, the intersection of their backward extensions, marks the position of the image. To check the accuracy of your drawing, draw a third ray through the center of curvature R.  (The ray or its extension must pass through R.) This ray reflects back onto itself.  The intersection of all the rays marks the image of the chosen point.

Example: concave mirror, real image

The image, in general, has not the same size of the object.  We define the magnification M as the ratio of the height of the image hi to the height of the object ho.  We have from geometry

M = hi/ho= -xi/xo

If the magnification is negative, the image is inverted.

Here are some things that always go together for curved mirrors.

 Real image <--> inverted image <--> xi is positive <--> M is negative Virtual image <--> upright image <--> xi is negative <--> M is positive

Convex mirrors form only virtual images.
Concave mirrors form real images, if xo > f and virtual images if xo < f.  The type of image formed depends on the position of the object.

Example: concave mirror, virtual image

Problem:

A concave mirror has a radius of curvature of 20 cm.  Find the location of the image for object distances (a) 40cm, (b) 20cm, (d) 10cm.
For each case state whether the image is real or virtual, and upright or inverted.  Find the magnification for each case.

Solution:
Mirror equation: 1/xo + 1/xi = 1/f
For a concave mirror f is positive.  f = R/2.  For this mirror f=10 cm.
 (a) 1/10 - 1/40 = 1/xi.  3/40 = 1/xi.  xi = 13.33 cm.  The image is real. M = -xi/xo = -1/3.  The image is inverted. (b) 1/10 - 1/20 = 1/xi.  1/20 = 1/xi.  xi = 20 cm.  The image is real. M = -xi/xo = -1.  The image is inverted. (c) 1/10 - 1/10 = 1/xi.  0 = 1/xi.  There is no finite image distance.  All rays from the object are parallel after reflection.

Example: convex mirror, virtual image

Problem:

A object 2 cm in height is placed 3 cm in front of a concave mirror.  If the image is 5 cm in height and virtual, what is the focal length of the mirror?

 Solution: M = -xi/xo = 5/2. xo = 3 cm, therefore xi = -7.5 cm. 1/xo + 1/x i= 1/f.  1/3 - 1/7.5 = 1/f.  f = 5 cm.

Problem:

A object 2 cm in height is placed 3 cm in front of a concave mirror.  If the image is 5 cm in height and virtual, what is the focal length of the mirror?

 Solution: M = -xi/xo = 5/2.  xo = 3 cm, therefore xi = -7.5 cm. 1/xo + 1/x i= 1/f.  1/3 - 1/7.5 = 1/f.  f = 5 cm.

Images can be formed reflection and by refraction.  Assume that we have a boundary between two media with different indices of refraction, and that the boundary has a radius of curvature R, as shown in the figure.  The center of curvature is at the origin.

Light is refracted at the boundary.  Two rays starting at an object are shown.  One ray is incident normally and is not refracted.  The other ray is incident at an angle θi with respect to the normal, and the refracted angle θt is given by Snell's law, nisinθi = ntsinθt.  An image is formed where the two rays intersect.

For small angles θ, (θ(rad) << 1) we have sinθ tanθ θ.  (See the graph on the right which plots θ(rad), sinθ, and tanθ versus θ(deg) for angles from 0 to 10o.)  This is called the small angle approximation.  It simplifies calculations if it is applicable.

Using the small angle approximation we can show that for a spherical interface

n1/xo + n2/xi = (n2 - n1)/R,

if we adopt the following sign conventions.

 The object's position xo is in front of the interface and the object distance is always positive. The image distance xi is positive if the image position lies behind the interface. The image distance xi is negative if the image position lies in front of the interface. The radius of curvature R is positive, if the center of curvature lies behind the interface. The radius of curvature R is negative, if the center of curvature lies in front of the interface.

The magnification is M = hi/ho = -(n1xi)/(n2xo).  If M is negative, the image is inverted.

Problem:

A simple model of the human eye ignores its lens entirely.  Much of what the eye does to light happens at the transparent cornea.  Assume that this outer surface has a radius of curvature of 6 mm, and assume that the eyeball contains just one fluid of index of refraction n = 1.4.  Prove that every distant object will be imaged on the retina, 21 mm behind the cornea.  Describe the image.

 Solution: For distant objects xo is very large and n1/xo is very small.  We can safely ignore this term in the equation n1/xo + n2/xi = (n2 - n1)/R.  We then have n2/xi = (n2 - n1)/R, which yields xi = n2R/(n2 - n1) = (1.4*6 mm)/0.4 = 21 mm.  The image distance is 21 mm.  The image is very small and inverted.

For a flat interface we have R = infinite and n1/xo + n2/xi = 0.
Let an object be underwater, 10 cm from the water-air interface.
We find xi = -(n2/n1)xo = -(1/1.33)xo.  xi = -7.5 cm.
The magnification is M = 1, it is positive.
The image is a virtual, non-inverted image.
A detector viewing the object from the air side sees the image at the apparent position closer to the interface.

Problem:

A cubical block of ice, 50 cm on a side, is placed on a level floor over a speck of dust.  Find the location of the image of the dust if the index of refraction of the ice is n = 1.309.

 Solution: xo = 50 cm.  xi = -(1/1.309)xo = -38.2 cm. The apparent position is 38.2 cm below the top surface of the ice.