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 straightline 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 straightline 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 x_{o} denote the perpendicular distance of the object from the mirror surface, and let x_{i} 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 x_{o }+ x_{i }= 0. For mirrors, negative image distances are associated with virtual images and positive image distances are associated with real images. 
For each ray θ_{i} = θ_{r}. 

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: 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 = 360^{o,} 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.



Curved MirrorsLight 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 x_{o}, then an image is formed
at an image distance x_{i}, where x_{o} and x_{i}
satisfy the mirror equation,
1/x_{o }+ 1/x_{i }= 1/f. Sign conventions for the mirror equation: x_{i} is positive for a real image in front of the mirror surface, and x_{i} is negative for a virtual image behind the mirror surface. x_{o} and x_{i} 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:

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 h_{i} to the height of
the object h_{o}. We have from geometry
M = h_{i}/h_{o}= x_{i}/x_{o}. If the magnification is negative, the image is inverted. Here are some things that always go together for curved mirrors.
Convex mirrors form only virtual images. 
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.


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 = x_{i}/x_{o }= 5/2. x_{o }= 3 cm, therefore x_{i }= 7.5 cm. 1/x_{o }+ 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 = x_{i}/x_{o }= 5/2. x_{o }= 3 cm, therefore x_{i }= 7.5 cm. 1/x_{o }+ 1/x_{ i}= 1/f. 1/3  1/7.5 = 1/f. f = 5 cm. 
Link: The Physics Classroom: Ray optics
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 θ_{i} with respect to the normal, and the refracted angle θ_{t} is given by Snell's law, n_{i}sinθ_{i }= n_{t}sinθ_{t}. 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 10^{o}.) 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 n_{1}/x_{o }+ n_{2}/x_{i }= (n_{2 } n_{1})/R, if we adopt the following sign conventions.
The magnification is M = h_{i}/h_{o }= (n_{1}x_{i})/(n_{2}x_{o}). 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.



For a flat interface we have R = infinite and n_{1}/x_{o
}+ n_{2}/x_{i }= 0. Let an object be underwater, 10 cm from the waterair interface. We find x_{i }= (n_{2}/n_{1})x_{o }= (1/1.33)x_{o}. x_{i }= 7.5 cm. The magnification is M = 1, it is positive. The image is a virtual, noninverted 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.


Please complete assignment 21 now.