91Ó°ÊÓ

The focal length of a lens is an important parameter that informs us about the lens's ability to bend light. It is the distance between the lens and the point where it converges or diverges light. For the calculation, we use the lens formula:

• \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)

In our exercise, we noted that the object distance \(d_o\) is 16.0 cm and the image distance \(d_i\) is -12.0 cm (since the image is on the same side as the object, it is considered negative). Thus, when we substitute these values into the lens formula, we find \(f\) to be -48.0 cm.
This negative focal length means that the lens is diverging. Diverging lenses spread light rays apart and are often used in applications like eyeglasses for nearsighted individuals.

A diverging lens, unlike a converging (convex) lens, causes parallel incoming light rays to spread apart upon exiting the lens. This is why it produces a negative focal length. Diverging lenses are characterized by their concave surfaces and always form virtual, upright, and smaller images compared to the object.

• They are used in applications like laser pointers, eyeglasses, and camera systems where dispersing light is needed.
• The focal point of a diverging lens is located on the same side as the incoming light, which means objects placed on that side will have images appearing to "spread out" from the focal point.

In practical terms, diverging lenses can be identified by their thin center compared to the edges, giving them their unique curved shape.

Lens magnification tells us how much larger or smaller an image is compared to the actual object. It is directly tied to the object and image distances given by the ratio:

• \(m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}\)

In the exercise, the lens magnification is calculated using object distance \(d_o\) and image distance \(d_i\). We find that the magnification is 0.75, indicating that the image is smaller than the object.
Knowing the object height \(h_o\) is 8.50 mm, the image height \(h_i\) computes to 6.375 mm.
This positive magnification indicates an upright image, which is expected of virtual images formed by diverging lenses.

Drawing a principal-ray diagram helps visualize how light travels through a lens and where images form. To draw one for a diverging lens:

• Begin by sketching the lens, marking the principal axis and the focal points.
• Draw a ray from the top of the object parallel to the principal axis, refracting it so it appears to diverge from the focal point on the same side after passing through the lens.
• Another key ray passes through the optical center of the lens without bending.

The intersection of these rays is where the virtual image forms. In our scenario, since the rays appear to extrapolate back from the focal point, the image is upright and smaller, confirming the calculations of the exercise.