Conceptual Physics

33.6 - Sample problem: object inside the focal point of a converging lens

The penguin is inside the focal point of a converging lens. What kind of image will result? Again, once you know the result, can you think of any

application that relies on a configuration like this?

Strategy

Use ray tracing to establish the nature of the image.

Diagram

Step-by-step solution

1. Ray 1 starts parallel to the principal axis. It refracts at the lens and passes through the focal point on the far side. To obtain a
   convergence point for the image, we extend the ray backward, showing this extension with a dashed line.


2. Ray 2 passes through the center of the lens without changing direction. We again extend a virtual ray backward.


3. Ray 3 must be horizontal on the far side of the lens and it must also pass through a focal point. Here, it starts at the top of the object,
   directed upward so that its backward extension passes through the focal point on the near side. It refracts at the lens, becoming
   horizontal on the far side. By extending the horizontal portion backward, we create a virtual ray that passes through the convergence
   point.

The result is an upright, virtual image that is larger than the object. This configuration is typical of objects being viewed in a magnifying glass.

The magnifying glass enlarges objects, making it easier to view their fine detail. The image cannot be projected, but it is upright, a matter of

great convenience to physicists and textbook authors over 40 who may use a magnifying glass to see fine details!

What kind of image will be produced?

33.7 - Lens equations

Ray diagrams provide a qualitative technique for establishing the nature of an image created by a thin lens. The image’s position and size can

be quantified with a set of three equations called the lens equations. These are the focus (we could not resist) of this section.

The thin lens equation is shown in Equation 1. It states the relationship between object distance, image distance, and focal length: The sum of

the reciprocals of the object distance and the image distance equals the reciprocal of the focal length.

For instance, suppose you want to calculate the location of an image created by placing an object 0.50 meters away from a lens with a focal

length of 0.33 meters. The thin lens equation and a little arithmetic reveal that the image distance is 1.0 meters. (We solve this problem in

Lens and magnification

equations

Interpretation of signs

(^612) Copyright 2007 Kinetic Books Co. Chapter 33