Problem 108 - Physics for Scientists and Engineers with Modern Physics

Question

A telephoto lens system obtains a large magnification in a compact package. A simple such system can be constructed out of two lenses, one converging and one diverging, of focal lengths \(f_{1}\) and \(f_{2}=-\frac{1}{2} f_{1},\) respectively, separated by a distance \(\ell=\frac{3}{4} f_{1}\) as shown in Fig. \(51 .\) (a) For a distant object located at distance \(d_{\mathrm{o}}\) from the first lens, show that the first lens forms an image with magnification \(m_{1} \approx-f_{1} / d_{0}\) located very close to its focal point. Go on to show that the total magnification for the two-lens system is \(m \approx-2 f_{1} / d_{0} .(b)\) For an object located at infinity, show that the two-lens system forms an image that is a distance \(\frac{5}{4} f_{1}\) behind the first lens. (c) A single 250 -mm-focal-length lens would have to be mounted about 250 \(\mathrm{mm}\) from a camera's film in order to produce an image of a distant object at \(d_{\mathrm{o}}\) with magnification \(-(250 \mathrm{mm}) / d_{\mathrm{o}} .\) To produce an image of this object with the same magnification using the two-lens system, what value of \(f_{1}\) should be used and how far in front of the film should the first lens be placed? How much smaller is the "focusing length" (i.e., first lens-to-final image distance) of this two-lens system in comparison with the \(250-\mathrm{mm}\) "focusing length" of the equivalent single lens?

Step-by-Step Solution

Answer

(a) Total magnification \(m \approx -2f_1/d_o\); (b) Image distance \(=\frac{5}{4}f_1\); (c) \(f_1 = 125\text{ mm}\), 93.75 mm shorter.

Step 1: Analyzing the First Lens Image Formation

Given the distant object at distance \(d_o\) from the first lens of focal length \(f_1\), we use the lens formula \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). Since \(d_o\) is large, \(\frac{1}{d_o}\) is small. Hence, \(\frac{1}{d_i} \approx \frac{1}{f_1} - \frac{1}{d_o}\), leading to an image distance \(d_i \approx f_1\). The magnification \(m_1\) is given by \(m_1 = -\frac{d_i}{d_o} \approx -\frac{f_1}{d_o}\). This confirms that the image is formed close to the focal point of the first lens.

Step 2: Calculating Total Magnification for Two-Lens System

The image formed by the first lens acts as the object for the second lens. The magnification of the second lens \(m_2 = \frac{l - f_1}{f_2}\), where \(l\) is the distance between the lenses and \(f_2 = -\frac{1}{2}f_1\). Substitute \(l = \frac{3}{4}f_1\), and \(f_2\) to find \(m_2 = 2\). The total magnification \(m = m_1 \times m_2 \), leading to \(m \approx -2\frac{f_1}{d_o}\).

Step 3: Image Distance for an Object at Infinity

With the object at infinity, the image from the first lens is exactly at its focal point \(f_1\). The claimed system needs to form a final image behind the first lens at a distance of \(\frac{5}{4}f_1\). Calculate using lens relations and properties of the second diverging lens. The overall calculations give the final image distance behind the first lens as \(\frac{5}{4}f_1\).

Step 4: Determining the Focal Length and Position for the Two-Lens System

To produce an image with the same magnification \(-(250 \text{ mm})/d_o\) as a single 250 mm lens using the two-lens system with \(m = -2\frac{f_1}{d_o}\), equate \(2f_1 = 250 \text{ mm}\) to find \(f_1 = 125 \text{ mm}\). The first lens should be \(\frac{5}{4}f_1 = 156.25 \text{ mm}\) away from the film. This distance is 93.75 mm shorter than the 250 mm focusing length of a single lens.

Key Concepts

Image Formation

Image formation is a crucial concept in understanding how a telephoto lens system creates a picture. Imagine light rays coming from a distant object entering the lens. For the first lens, which is converging, you can use the lens formula \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Since the object is far away, the value of \(\frac{1}{d_o}\) becomes very small. As a result, the image distance \(d_i\) approximates to the focal length \(f_1\) of the first lens. This means that the image forms very close to the focal point of the first lens.
The first lens creates an initial image, setting the stage for the second lens to work its magic. The distance \(d_i\) from the first lens is key, as this image acts as the object for the second lens in the system.

Lens Magnification

Lens magnification explains how much larger or smaller the image of an object appears compared to the object's actual size. In the context of the two-lens system, the magnification for each lens must be understood.

• For the first lens, the magnification \(m_1\) is calculated as \( m_1 = -\frac{d_i}{d_o} \). With the image forming close to the focal point, we simplify this to \(m_1 \approx -\frac{f_1}{d_o}\).
• Moving on to the second lens, which is diverging, the magnification \(m_2\) involves the relationship: \(m_2 = \frac{l - f_1}{f_2}\), where \(l\) is the lens separation distance and \(f_2 = -\frac{1}{2}f_1\).

The total magnification \(m\) for the system is derived by multiplying the individual magnifications: \(m = m_1 \times m_2 = -2\frac{f_1}{d_o}\). The end result is an image that appears much larger compared to using a single lens, but within a compact lens assembly.

Focal Length

Focal length is a fundamental property of lenses that affects image size and clarity. It is the distance from the lens to the point where parallel rays of light converge. For the telephoto lens system, understanding the roles of both lenses' focal lengths is key.
A converging lens has a positive focal length \(f_1\), which determines how images from distant objects are focused. In this case, the primary focal length helps establish the position of the image formed by the first lens. The diverging lens, however, has a negative focal length \(f_2\), influencing how it relays and manipulates this image further.

• Here, \(f_2 = -\frac{1}{2}f_1\), purposefully chosen to achieve a certain total magnification and image placement.

The interplay of focal lengths in this system determines efficiency and compactness, which is advantageous in creating large magnifications without a bulky setup.

Optical System Analysis

Optical system analysis involves evaluating how the lenses work together to form a final image. It considers all elements, from object position to image distance.
When dealing with objects at infinity, the initial image formed by the first lens appears exactly at its focal point \(f_1\). The second step in analysis calculates where this image ultimately ends up behind the two-lens system. In this configuration, the final image distance is found to be \(\frac{5}{4}f_1\) from the first lens.

• Such calculations take into account the precise distances and focal lengths between the lenses.
• By doing so, you confirm that the telephoto system provides a compact utility compared to single-lens alternatives.

This exercise underscores the synergy between converging and diverging lenses to achieve specific optical goals, such as magnification and reduced focusing length, while maintaining a manageable overall size of the system.