91Ó°ÊÓ

The lens equation is a fundamental formula used in optics, helping us understand how converging lenses, like the one in this exercise, form images. It is represented by the equation \( \frac{1}{f} = \frac{1}{d_0} + \frac{1}{d} \), where:

• \( f \) is the focal length of the lens,
• \( d_0 \) is the distance from the lens to the object, and
• \( d \) is the distance from the lens to the image.

Knowing how to rearrange and use this equation can help solve many problems related to lenses. For instance, if the desired image distance \( d \) and the lens's focal length \( f \) are known, you can rearrange the equation to find the object distance \( d_0 \). In our case, we needed the image to form at \( 3f \). Substituting this into the lens equation allowed us to solve for \( d_0 \), leading to \( d_0 = \frac{3f}{2} \). This indicates how far the object should be placed from the lens to achieve the required image distance.

Magnification tells us how much larger or smaller the image is compared to the object. It also gives us information about the orientation of the image. In optics, magnification \( M \) can be calculated using the formula: \( M = \frac{-d}{d_0} \), where:

• \( d \) is the image distance,
• \( d_0 \) is the object distance, and
• The negative sign indicates that a negative magnification means the image is inverted.

In our exercise, with \( d = 3f \) and \( d_0 = \frac{3f}{2} \), plugging these into the magnification formula gives \( M = -2 \). This result tells us two things:

• The image is flipped upside down as the magnification is negative, and
• It is twice the size of the original object.

Image formation through lenses is a key concept in understanding how lenses function in various applications, from eyeglasses to cameras. A converging lens, as used in our problem, bends light rays inward, converging them at a point to form an image.
Images formed by lenses can be either real or virtual. A real image is formed when the light actually converges at the image location and can be projected onto a screen. Conversely, a virtual image forms when the light rays do not actually meet but only appear to converge.
In the given exercise, the image formed at a distance \( 3f \) on the right side of the lens is real and inverted, which is confirmed by the negative magnification value. Understanding how the arrangement of the object, lens, and placement affects the type and orientation of the image is crucial for effectively utilizing lenses in practical scenarios.