Problem 113 - University Physics with Modern Physics

Question

Focal Length of a Zoom Lens. Figure P34.113 shows a simple version of a zoom lens. The converging lens has focal length \(f_{1},\) and the diverging lens has focal length \(f_{2}=-\left|f_{2}\right|\) The two lenses are separated by a variable distance \(d\) that is always less than \(f_{1} .\) Also, the magnitude of the focal length of the diverging lens satisfies the inequality \(\left|f_{2}\right|>\left(f_{1}-d\right) .\) To determine the effective focal length of the combination lens, consider a bundle of parallel rays of radius \(r_{0}\) entering the converging lens. (a) Show that the radius of the ray bundle decreases to \(r_{0}^{\prime}=r_{0}\left(f_{1}-d\right) / f_{1}\) at the point that it enters the diverging lens. (b) Show that the final image \(I^{\prime}\) is formed a distance \(s_{2}^{\prime}=\left|f_{2}\right|\left(f_{1}-d\right) /\left(\left|f_{2}\right|-f_{1}+d\right)\) to the right of the diverging lens. (c) If the rays that emerge from the diverging lens and reach the final image point are extended backward to the left of the diverging lens, they will eventually expand to the original radius \(r_{0}\) at some point \(Q .\) The distance from the final image \(I^{\prime}\) to the point \(Q\) is the effective focal length \(f\) of the lens combination; if the combination were replaced by a single lens of focal length \(f\) placed at \(Q,\) parallel rays would still be brought to a focus at \(I^{\prime}\) . Show that the effective focal length is given by \(f=f_{1}\left|f_{2}\right| /\left(\left|f_{2}\right|-f_{1}+d\right) .\) (d) If \(f_{1}=12.0 \mathrm{cm}\) \(f_{2}=-18.0 \mathrm{cm},\) and the separation \(d\) is adjustable between 0 and \(4.0 \mathrm{cm},\) find the maximum and minimum focal lengths of the combination. What value of \(d\) gives \(f=30.0 \mathrm{cm} ?\)

Step-by-Step Solution

Answer

Maximum focal length: 36 cm; minimum focal length: 18 cm; \(d\) for \(f=30\): 1.2 cm.

Step 1: Understand the First Lens Effect

When parallel rays enter the first lens (converging lens of focal length \(f_1\)), they start converging towards its focal point. The distance from where these rays focus (if the diverging lens weren't placed) is \(f_1\). However, the diverging lens is placed before this happens at a distance \(d

Step 2: Calculate the Radius Before Entering Diverging Lens

Between the converging lens and the diverging lens, the radius of the light bundle changes. The radius decreases linearly since the lens attempts to focus the rays. The decrease is proportional to the fraction of the focal length at which the rays are. Thus, the new radius at the diverging lens \(r_0'\) is computed as \(r_0' = r_0 \left(\frac{f_1-d}{f_1}\right)\).

Step 3: Apply Lens Formula to Diverging Lens

For the diverging lens, use the lens maker's equation in the new setup. The image from the converging lens acts as a virtual object for the diverging lens. Applying lens formula \(\frac{1}{f_2} = \frac{1}{v_2} - \frac{1}{u_2}\) where \(f_2\) is negative, will show that the distance to the final image formed by the diverging lens \(s_2'\) is given by the formula \(s_2' = \frac{|f_2|(f_1-d)}{|f_2| - (f_1-d)}\).

Step 4: Calculate Effective Focal Length

The point \(Q\) is where the diverging rays, if extended backward, match the original radius \(r_0\). Therefore, using symmetry of the rays entering and exiting, the effective focal length \(f\) is derived to be \(f = \frac{f_1 |f_2|}{|f_2| - f_1 + d}\).

Step 5: Plug in Values to Determine Maximum and Minimum Focal Lengths

Substitute \(f_1 = 12.0\,\text{cm}\), \(f_2 = -18.0\,\text{cm}\), and explore the range of \(d\) from 0 to 4 cm. Compute the effective focal length \(f\) for \(d = 0\) and \(d = 4\,\text{cm}\) to find the maximum and minimum focal lengths. Compute \(d\) such that \(f = 30.0\,\text{cm}\).

Step 6: Calculations

At \(d = 0\), \(f = \frac{12 \times 18}{18 - 12} = 36\,\text{cm}\). At \(d = 4\,\text{cm}\), \(f = \frac{12 \times 18}{18 - 12 + 4} = 18\,\text{cm}\). To find \(d\) for \(f = 30\,\text{cm}\), solve \(\frac{12 \times 18}{18 - 12 + d} = 30\), giving \(d = 1.2\,\text{cm}\).

Key Concepts

Focal Length Calculation

When working with a zoom lens that involves two lenses - a converging lens and a diverging lens - understanding how to calculate the resulting focal length is crucial. The focal length of a lens is the distance between the lens and the point where it focuses light.
For a combined lens system, like a zoom lens, the effective focal length tells us where light rays from a distant point will focus after passing through the entire system.

In this specific problem, the focal length of the converging lens is denoted by \(f_1\), and for the diverging lens, it's \(f_2 = -\left|f_2\right|\). These two lenses are separated by a distance \(d\) which can be adjusted. By carefully studying how they interact with each other, we can determine how the system as a whole behaves optically.
Using the formula given in the problem, the effective focal length \(f\) of this combination is calculated by:

• \[ f = \frac{f_1 \left| f_2 \right|}{\left| f_2 \right| - f_1 + d} \]

This mathematical expression helps us understand how the focal lengths of the individual lenses and the distance \(d\) between them affects the overall focal length of the lens system.

Converging and Diverging Lenses

Converging lenses, often referred to as positive lenses, are designed to focus light. When parallel light rays pass through a converging lens, they bend towards each other, meeting at the lens's focal point.
This is why the focal length for a converging lens like the one given by \(f_1\) is positive.

Diverging lenses, in contrast, spread out light rays. They are known as negative or concave lenses. When light passes through a diverging lens, the rays diverge and appear to originate from a focal point behind the lens. This results in a negative focal length, represented here as \(f_2 = -\left|f_2\right|\).
This dual lens system works together to adjust the focus and magnification of an optical device. Understanding how these lenses work in conjunction helps us apply this knowledge practically, such as when designing zoom lenses in cameras. The converging lens gathers the incoming light, while the diverging lens adjusts the focus and spreads it according to the needed focal length of the entire system.

Lens Equation Application

To accurately determine how a lens system will behave, the lens equation is a vital tool. The lens equation mathematically relates the object distance, image distance, and focal length. It is expressed as:

• \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)

where \(f\) is the focal length, \(v\) is the image distance, and \(u\) is the object distance from the lens.
When applying this to the problem at hand, after parallel rays pass through the converging lens, they become the object for the diverging lens.
By using the modified lens equation, you can calculate the position \(s_2'\) where the rays will focus after passing through both lenses:

• \[ s_2' = \frac{|f_2|(f_1-d)}{|f_2| - (f_1-d)} \]

This equation helps predict the final image location relative to the diverging lens. The step-by-step calculations give insight into how modifications in the distance between the lenses or changes in either lens's focal length will influence where an image is formed. Application of these formulae shows how engineers and scientists achieve desired imaging results in optical devices.

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Key Concepts

Focal Length Calculation

When working with a zoom lens that involves two lenses - a converging lens and a diverging lens - understanding how to calculate the resulting focal length is crucial. The focal length of a lens is the distance between the lens and the point where it focuses light.
For a combined lens system, like a zoom lens, the effective focal length tells us where light rays from a distant point will focus after passing through the entire system.

In this specific problem, the focal length of the converging lens is denoted by \(f_1\), and for the diverging lens, it's \(f_2 = -\left|f_2\right|\). These two lenses are separated by a distance \(d\) which can be adjusted. By carefully studying how they interact with each other, we can determine how the system as a whole behaves optically.
Using the formula given in the problem, the effective focal length \(f\) of this combination is calculated by:

• \[ f = \frac{f_1 \left| f_2 \right|}{\left| f_2 \right| - f_1 + d} \]

This mathematical expression helps us understand how the focal lengths of the individual lenses and the distance \(d\) between them affects the overall focal length of the lens system.

Converging and Diverging Lenses

Converging lenses, often referred to as positive lenses, are designed to focus light. When parallel light rays pass through a converging lens, they bend towards each other, meeting at the lens's focal point.
This is why the focal length for a converging lens like the one given by \(f_1\) is positive.

Diverging lenses, in contrast, spread out light rays. They are known as negative or concave lenses. When light passes through a diverging lens, the rays diverge and appear to originate from a focal point behind the lens. This results in a negative focal length, represented here as \(f_2 = -\left|f_2\right|\).
This dual lens system works together to adjust the focus and magnification of an optical device. Understanding how these lenses work in conjunction helps us apply this knowledge practically, such as when designing zoom lenses in cameras. The converging lens gathers the incoming light, while the diverging lens adjusts the focus and spreads it according to the needed focal length of the entire system.

Lens Equation Application

To accurately determine how a lens system will behave, the lens equation is a vital tool. The lens equation mathematically relates the object distance, image distance, and focal length. It is expressed as:

• \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)

where \(f\) is the focal length, \(v\) is the image distance, and \(u\) is the object distance from the lens.
When applying this to the problem at hand, after parallel rays pass through the converging lens, they become the object for the diverging lens.
By using the modified lens equation, you can calculate the position \(s_2'\) where the rays will focus after passing through both lenses:

• \[ s_2' = \frac{|f_2|(f_1-d)}{|f_2| - (f_1-d)} \]

This equation helps predict the final image location relative to the diverging lens. The step-by-step calculations give insight into how modifications in the distance between the lenses or changes in either lens's focal length will influence where an image is formed. Application of these formulae shows how engineers and scientists achieve desired imaging results in optical devices.