Problem 77 - University Physics with Modern Physics

Question

(a) You want to use a lens with a focal length of 35.0 cm to produce a real image of an object, with the height of the image twice the height of the object. What kind of lens do you need, and where should the object be placed? (b) Suppose you want a virtual image of the same object, with the same magnification-what kind of lens do you need, and where should the object be placed?

Step-by-Step Solution

Answer

(a) Use a converging lens, place object 70 cm from lens. (b) Use a diverging lens, place object 17.5 cm from lens.

Step 1: Understanding the Problem

We need to determine the type of lens and the object distance to produce first a real image with a specific magnification, and then a virtual image with the same magnification.

Step 2a: Real Image with Magnification

For a real image where the image height is twice the object height, the magnification \(m\) is \(-2\) (image is inverted in real scenarios). Magnification is defined as \(m = \frac{-d_i}{d_o}\), where \(d_i\) is the image distance and \(d_o\) is the object distance.

Step 3a: Lens Formula for Real Image

Use the lens formula \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where \(f = 35.0\, \text{cm}\). Substitute \(d_i = -2d_o\) into the formula to find \(d_o\).

Step 4a: Solving for Object Distance (Real Image)

From \(\frac{1}{35} = \frac{1}{d_o} + \frac{1}{-2d_o}\), we rewrite as \(\frac{1}{35} = \frac{1}{d_o} - \frac{1}{2d_o}\). Simplifying gives: \(\frac{1}{35} = \frac{1}{2d_o}\). Therefore, \(d_o = 70\, \text{cm}\). Place the object 70 cm from the lens.

Step 5a: Type of Lens for Real Image

The lens needs to be a converging lens (positive focal length) to produce a real image with the object 70 cm away.

Step 2b: Virtual Image with Magnification

For a virtual image with the same magnification (twice the object's height), the magnification \(m\) is \(+2\).

Step 3b: Lens Formula for Virtual Image

Use \(d_i = -m \cdot d_o = -2d_o\) (since it's a virtual image). Substitute into the lens formula to solve for \(d_o\).

Step 4b: Solving for Object Distance (Virtual Image)

From \(\frac{1}{35} = \frac{1}{d_o} - \frac{1}{2d_o}\), we rearrange to \(\frac{1}{35} = -\frac{1}{2d_o}\). Solving gives \(d_o = -17.5\, \text{cm}\). Hence, the object should be placed 17.5 cm in front of the lens.

Step 5b: Type of Lens for Virtual Image

The lens for a virtual image must be a diverging lens (negative focal length) when using the same setup.

Key Concepts

Focal Length

The focal length of a lens is the distance between its center and the point where it brings parallel rays (such as those from a distant object) into focus. It is a critical parameter in lens design as it influences image size and type. A positive focal length indicates a converging lens, which can focus light to a point. Conversely, a negative focal length represents a diverging lens that spreads light rays apart. For instance, in the problem, you are given a focal length of 35.0 cm to determine the optimal lens setup for producing both real and virtual images. Knowing whether this focal length is positive or negative will guide you in choosing the correct type of lens.

Real Image

A real image is formed when light rays actually converge at a point after passing through a lens or reflecting off a mirror. These images can be projected onto a screen, as they exist physically. Real images are usually inverted compared to the original object. In the exercise, when needing a real image with magnification of -2, the converging lens is necessary. The lens ensures that light rays from the object converge to form the image 70 cm away, which is double the object size and inverted.

Virtual Image

Virtual images are those where the light rays appear to originate but do not actually meet. These images cannot be projected onto a screen as they rely on the perception of the observer. Virtual images are upright due to the manner in which the lens bends the light rays. To achieve a virtual image with a magnification of +2, the setup in the exercise uses a diverging lens, where the object is placed closer to the lens at 17.5 cm. Here, the lens creates an image that seems to be larger than, but in a different position than, the actual object.

Object Distance

The object distance, denoted as \(d_o\), is the distance between the object and the lens. It is a crucial measurement as it influences both the type and the size of the image formed. In conjunction with the lens formula \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where \(d_i\) is the image distance, you can determine the necessary positioning of the object relative to the lens to achieve the desired image magnification. In the problem, for a real image, placing the object 70 cm from the lens fits the conditions of the formula. For a virtual image, placing it at 17.5 cm alters the conditions to produce the virtual image effect with the same magnification but different orientation.

Converging Lens

A converging lens, often called a convex lens, is thicker at the center than at its edges. It bends light rays toward each other, bringing parallel rays to a focal point. These lenses are frequently used in applications where light needs to be focused, such as in cameras, glasses, and telescopes. In this exercise, to obtain a real, inverted image with a magnification of -2, the converging lens with a 35 cm focal length is used. It compels the rays to meet again at the image position, thereby enlarging the image to twice its original size and successfully flipping it.

Diverging Lens

Diverging lenses, or concave lenses, are thinner at the center than at the edges and cause light rays to spread out. These lenses are useful in correcting vision issues like myopia (nearsightedness) and are used in various visual devices. For the exercise's virtual image requirement, a diverging lens serves to manipulate light rays in such a way that they indicate a larger, upright, virtual image when the object is positioned 17.5 cm from the lens. The negative focal length inherent to diverging lenses plays an essential role in creating the conditions necessary for the image to appear twice the size of the object.