Chapter 23
The Giancoli onLine Tutor

Light: Geometric Optics

I. Mirrors

I-a. Concave Mirrors

Much of this chapter is devoted to solving lens and mirror problems using the equation:

do is the distance from the lens of the object, di is resulting distance of the image and f is the focal length of the lens or mirror. Although it is possible to have an object to the right of the lens (a virtual object), our objects will always be to the left of the lens and the object distance will always be positive. Concave mirrors, , have a positive focal length, while Convex mirrors, , have a negative focal length. You can solve the equation above by simple substitution. Positive image distances are to the left of the mirror and those that are negative are to the right of the mirror.

Interactive Example

Suppose f = 10 cm and do is 20 cm, what is di? Enter your answer and click the clover for good luck (and the answer) . 



Ray Tracing

It is wise to check your answer by drawing a simple diagram. The blue line represents a ray of light coming from the top of the object arrow and moving parallel, such a ray will be reflected through the (violet) focal point. The red ray goes through the focal point and is reflected parallel. The green line strikes the middle of the mirror and is reflected at an equal angle. Where they cross is where the image is located. As you see the image is where you predicted! Cool!

In this case the image is upside down (inverted) and the same size as the object and that is because the magnification is given by

m = -di/do


In this case -1. The minus sign means it is inverted and the '1' means there is no magnification, either way.

Interactive Example: Practice->


You will have noticed that some of the image distances in the above Example are negative. These correspond to virtual images; images that are not actually formed but can be viewed, as shown in the picture on the left.

I-b. Convex Mirrors

Convex mirrors have a negative focal point and always produce virtual images.

Ray Tracing

  1. A ray coming in parallel to the axis will be reflected as if it was coming from the focal point.
  2. A ray heading for the center of curvature will reflect directly back on itself.
  3. A ray striking the mirror at its center will be reflected at an equal angle.
The eye extrapolates these backwards and where they appear to cross is the location of the virtual image.



Interactive Example: Practice->

II. Lenses

II-a. Converging Lenses

Converging lenses, , are thicker in the middle than at the edges.
In dealing with lenses you use the same equation as used for mirrors. The only differences are a positive image distance (real image) is on the right side of the lens. A negative image distance is a (virtual) image formed on the same side as the object. Also lenses have two focal points.

Ray Tracing

Simple ray tracing can check your answer. The red ray coming in parallel from the left goes out through the focal point on the right. The blue ray goes through the focal point on the left and comes out parallel on the right. The green ray striking the middle of the lens goes through undeflected.

Interactive Example: Practice->



II-b. Diverging Lenses

Diverging lenses, are thicker at the edges. As with the converging lens you use the 'mirror' equation subject to the new sign conventions outlined there.

Ray Tracing

  1. A ray coming in parallel comes out as if it is coming from the focal point.
  2. A ray striking the lens at its center goes through undeflected.
Where the rays cross is where the virtual image appears to be.

Interactive Example: Practice->


III. Lens Dynamics

III-a Index of Refraction

You need Quicktime 3 or better to use this Activity.

Interactive Activity

Fermat's Principle states that light travels from point 1 to point 2 in the least possible time. This gives rise to snell's law of refraction for light:
n1sin(q1)= n2sin(q2)
where q1 is the angle of incidence (with the perpendicular) at a surface , q2 is the angle of refraction in the material, such as, glass, on the other side of the surface, n1 and n2 are the indices of refraction above and below the surface, respectively. For light the index of refraction is c/v where c is the speed of light in a vacuum and v is the speed in the material in question. Actually this law applies to all waves going from 1 medium to another where the speed changes, such as, water waves entering shallow water. The application below is designed to help you understand this.

You are runner whose task it is to run from point 1 to point 2 in the shortest possible time. The gray area is concrete on which you can run at 9 m/s and the brown area is soft sand on which you can run at 5.2 m/s.
Click and drag the transparent protractor to the point at which you think you should cross the border and release the mouse button. Click on the time button to find your transit time in seconds. Readjust your crossing point until you find the minimum time which is 52 seconds. When you find it, Feedback will be given in the Text Area below.
Transit Time:
FeedBack:


As illustrated above light going into a region of higher index of refraction (lower speed) bends towards the normal (perpendicular) to the surface. If it going into a region of lower index (n1) it bends away from the normal. The equation that governs this is Snell's Law.

Interactive Example

If n1 = 1, q1 = 37o, q2 = 30o to the nearest two decimal places what is n2? Enter your answer and click the clover for good luck .
 

III-b. Internal Reflection

Another type of a problem deals with total internal reflection. This occurs when Snell's Law dictates that the refracted ray is at or greater than 90o. The critical angle, qc, is defined when q2 is 90o:

sinqc= n2/n1


If n1=1.5 and n2 = 1 the critical angle (to the nearest degree) is
Enter your answer and click the clover for good luck .
 

III-c. The Lens Makers Equation

The final type of problem is the lens makers equation.

1/f = (n-1)(1/R1+1/R2)


where n is the index refraction of the glass or plastic lens material, R1 is the radius of curvature of the first surface and R2 that of the second. f is then the focal length of the lens. The trick here is to keep track of the correct sign for R. If the surfaces curve out, as shown in this figure the R's are positive. If either surface curves in, that radius is considered to be negative.

n = 1.5, R1 = 10 cm and R2 = -5 cm, what is the focal length to the nearest cm?
Enter your answer and click the clover for good luck .
 

finis © 1999 Carl Adler mailto:Carl@Image-ination.com