91Ó°ÊÓ

When dealing with lenses in geometric optics, the lens formula is crucial for determining the image position created by a lens. This formula is given by \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]where:

• \( f \) is the focal length of the lens
• \( v \) is the image distance from the lens
• \( u \) is the object distance from the lens

A negative object distance means the object is on the same side as the incoming light, which is the case in this exercise with a given object distance of \(-20\, \text{cm}\). Using the first lens with \( f = 10\, \text{cm} \), the image distance \( v \) was calculated as \( 20\, \text{cm} \), meaning the image is formed on the opposite side of the lens to the object. This process is repeated for the second lens using its respective focal length, showing how each lens behaves as an individual optical element in the system.

Ray tracing is a graphical method used in geometric optics to determine the path of light as it travels through optical systems like lenses. It helps visualize how images are formed and where they appear. For the given exercise, drawing is done on a scaled grid or with a ruler. Here are some basic steps:

• Draw a parallel ray from the top of the object to the lens and refract it through the focal point on the opposite side.


• Pass a ray through the center of the lens straight, as it won’t bend.


• The intersection of these refracted rays gives the image position.

For the first lens, this process places the image at \(20\, \text{cm}\) on the other side, consistent with calculations using the lens formula. Repeating these steps for the second lens, considering the image from the first as the new object, confirms both the image location and height calculated earlier. Ray tracing offers a method to verify solutions visually, as was done accurately in this exercise.

Magnification in geometric optics tells us the size of the image relative to the object's size and whether it is inverted or upright. The magnification \( m \) is calculated using:\[ m = -\frac{v}{u} = \frac{h'_i}{h_o} \]where:

• \( h'_i \) is the height of the image
• \( h_o \) is the height of the object

For the first lens, substituting the calculated \( v \) and \( u \), we find the image is \(-1.0\, \text{cm}\) tall, indicating it is inverted and half the original object's height. Further calculations with the second lens, based on its image distance, yield a final image height of \(1.5\, \text{cm}\). This positive result denotes an upright image and reflects how these equations relate image size and orientation. The magnification equation provides insight into image characteristics, forming a key part of understanding optics.