Problem 16 - An Introduction to Physical Science

Question

An object is placed in front of a converging lens at an object distance of twice the focal length of the lens. Sketch a ray diagram and compare the image and object distances. Repeat with two more ray diagrams, using different focal lengths and still making the object distance twice the focal length. Can you draw any conclusions by comparing the object distance and the image distance?

Step-by-Step Solution

Answer

The image distance equals the object distance, both twice the focal length.

Step 1: Sketching the Ray Diagram

To begin, sketch a converging lens and mark its optical center. Place the object at a position that is twice the focal length ( 2f ) from the lens on the left side. Draw three principal rays from the top of the object: (1) a ray parallel to the principal axis, which refracts through the focal point on the opposite side, (2) a ray through the focal point in front of the lens, which refracts parallel to the principal axis, and (3) a ray through the optical center of the lens, which continues in a straight line. The point where these rays converge on the other side of the lens indicates the top of the image.

Step 2: Calculating Image Distance with First Focal Length (f)

Using the lens formula, \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where \(d_o = 2f\). Substitute this value: \(\frac{1}{f} = \frac{1}{2f} + \frac{1}{d_i}\). Solve for \(d_i\) to find that the image distance is also \(2f\).

Step 3: Re-sketching with a Second Focal Length

Repeat the sketch using a different focal length, say \(f' = 0.5f\). With this setup, the object distance \(d_o = 2f' = f\). Recalculate the image distance using the lens formula, where \(\frac{1}{f'} = \frac{1}{f} + \frac{1}{d_i}\). Solve for \(d_i\) to confirm the calculation, which remains \(2f'\).

Step 4: Re-sketching with a Third Focal Length

Change the focal length again, maybe \(f'' = 1.5f\). The object distance becomes \(d_o = 2f'' = 3f\). Use the lens formula to solve: \(\frac{1}{f''} = \frac{1}{3f} + \frac{1}{d_i}\). Calculate \(d_i\), and again, you will find that the image distance is \(2f''\).

Step 5: Conclusion from Comparing Distances

Upon reviewing the results for each focal length scenario, you will notice a pattern that the image distance \(d_i\) remains equal to \(d_o\) for an object distance of twice the focal length, irrelevant of the lens's focal length value. Therefore, whenever an object is placed at twice the focal length of a converging lens, the image distance is equidistant on the opposite side.

Key Concepts

Converging Lens

A converging lens is designed to bend incoming parallel light rays so they meet, or converge, at a single focal point. These lenses are typically thicker at the center than at the edges. This unique shape is what directs parallel rays to focus. Converging lenses are commonly used in various optical devices, including cameras, microscopes, and eyeglasses.

The primary function of a converging lens is to create a real and inverted image of an object placed at a distance. It achieves this by refracting light in such a way that all incoming rays aiming at the optical center are bent towards the principal axis, making them converge at the focal plane on the opposite side.

• The point where light rays converge is known as the focal point.
• The distance from the lens to the focal point is called the focal length.
• Converging lenses can create real and virtual images based on the object's position.

Ray Diagram

A ray diagram is a visual tool used to trace the path that light rays take as they pass through a lens. It's like a map showing how light behaves. When constructing a ray diagram for a converging lens, there are standard rays to consider.

Constructing Ray Diagrams

The three principal rays used in these diagrams include:

• Ray 1: Travels parallel to the principal axis and refracts through the focal point on the opposite side.
• Ray 2: Travels through the focal point in front of the lens and refracts parallel to the principal axis.
• Ray 3: Passes directly through the optical center of the lens without changing direction.

The point where these rays intersect on the opposite side of the lens represents the top of the image. Visualizing these paths helps understand how the object translates into an image through the lens.

Focal Length

The focal length of a lens is a crucial parameter in optics. It is the distance between the lens and the focal point where converged rays meet. In a converging lens, the focal point is located on the side of the lens opposite to the object.

The focal length determines several characteristics of the lens:

• Shorter focal lengths result in a stronger lens that bends light more sharply.
• Longer focal lengths create a gentler convergence, resulting in a different magnification rate.

In many scenarios, including the exercise, the relation between the object distance, image distance, and focal length can be explored using the lens formula:

\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}.\]Understanding focal length helps predict the behavior of lenses in optical systems, an essential skill in physics and engineering.

Image Distance

Image distance is the space between the lens and the formed image. In the case of a converging lens, if an object is placed at twice the focal length from the lens, the image is also formed at twice the focal length on the opposite side. This concept is consistently observed regardless of the variations in the lens's focal length.

Lens Formula and Image Distance

The lens formula \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) is instrumental in determining image distance \(d_i\) for any given object distance \(d_o\) and focal length \(f\). For an object positioned at twice the focal length, when substituting \(d_o = 2f\), it results in \(d_i\) being equal to \(d_o\). This indicates that the image distance mirrors the object distance in such scenarios.Overall, understanding image distance helps predict where and how an image will form through a lens, which is vital in designing optical devices. Having knowledge of this relationship enhances understanding of broader optical principles.