Question

A 1.20-cm-tall object is 50.0 cm to the left of a converging lens of focal length 40.0 cm. A second converging lens, this one having a focal length of 60.0 cm, is located 300.0 cm to the right of the first lens along the same optic axis. (a) Find the location and height of the image (call it \(I_1\)) formed by the lens with a focal length of 40.0 cm. (b) \(I_1\) is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses.

Short answer

The final image is located 76.5 cm to the right of the second lens with a height of -0.147 cm.

Step-by-step solution

Understand the Lens Formula

For a lens, the relationship between the focal length \(f\), the object distance \(d_o\), and the image distance \(d_i\) is given by \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\] We will use this formula to find the image location for each lens.

Calculate Image Location for the First Lens

The object is 50.0 cm to the left of the first lens, so the object distance \(d_o = -50.0\) cm (negative because the object is real and on the left). The focal length of the lens \(f = 40.0\) cm. Using the lens formula:\[ \frac{1}{40.0} = \frac{1}{-50.0} + \frac{1}{d_i} \]Solving for \(d_i\), we find:\[ \frac{1}{d_i} = \frac{1}{40} + \frac{1}{50} = \frac{5}{200} + \frac{4}{200} = \frac{9}{200} \]Thus, \(d_i = \frac{200}{9} \approx 22.22\) cm. This means the image \(I_1\) is 22.22 cm to the right of the first lens.

Calculate Image Height for the First Lens

The magnification \(m\) is given by:\[ m = -\frac{d_i}{d_o} \]Substitute the values:\[ m = -\frac{22.22}{-50.0} = 0.444 \]The height of the image \(h_i\) is given by:\[ h_i = m \times h_o = 0.444 \times 1.20 \approx 0.533 \text{ cm}\]Thus, the image \(I_1\) has a height of approximately 0.533 cm.

Determine Object Distance for the Second Lens

The image \(I_1\) formed by the first lens serves as the object for the second lens. Since the image is 22.22 cm to the right of the first lens and the second lens is 300 cm to the right of the first lens, the object distance for the second lens \(d_{o2} = 300.0 - 22.22 = 277.78\) cm to the left of the second lens.

Calculate Image Location for the Second Lens

The focal length for the second lens \(f = 60.0\) cm. Using the lens formula for the second lens:\[ \frac{1}{60.0} = \frac{1}{277.78} + \frac{1}{d_{i2}} \]Solving for \(d_{i2}\), we get:\[ \frac{1}{d_{i2}} = \frac{1}{60} - \frac{1}{277.78} = \frac{277.78 - 60}{60 \times 277.78} \approx \frac{217.78}{16666.8} \]\[ d_{i2} \approx 76.5 \text{ cm} \]The image is located approximately 76.5 cm to the right of the second lens.

Calculate Image Height for the Second Lens

Use magnification for the second lens:\[ m_2 = -\frac{d_{i2}}{d_{o2}} = -\frac{76.5}{277.78}\approx -0.275 \]The height of the final image \(h_{i2}\) is:\[ h_{i2} = m_2 \times h_{i1} = -0.275 \times 0.533 \approx -0.147 \text{ cm} \]The negative sign indicates the final image is inverted with respect to the object \(I_1\).

Key concepts

Lens Formula

The lens formula is essential in optical systems as it connects three core quantities: the focal length (\( f \)), the object distance (\( d_o \)), and the image distance (\( d_i \)). This relationship is expressed mathematically as \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]. Here’s a brief rundown of the meaning:

• Focal Length (\( f \)): This is a constant that represents the distance over which initially parallel rays converge (or appear to converge) after passing through the lens.
• Object Distance (\( d_o \)): It’s the distance from the object to the lens. By convention, it is negative if the object is in front of the lens.
• Image Distance (\( d_i \)): This is the distance from the lens to the image. It is positive if the image is on the opposite side of the object relative to the lens.

Using the formula helps in determining where an image will form, which is fundamental when working with multiple lenses in systems like the one described in our exercise.

Image Distance Calculation

Calculating the image distance involves rearranging the lens formula to solve for \( d_i \). Let's look at its application step by step.For the first lens:1. The object is 50 cm to the left of the lens, so \( d_o = -50.0 \) cm (since it’s a real object).2. The focal length is given as \( f = 40.0 \) cm.3. Plugging the values into the lens formula: \( \frac{1}{40.0} = \frac{1}{-50.0} + \frac{1}{d_i} \).Solving for \( d_i \), we have:- Add fractions: \( \frac{1}{d_i} = \frac{1/40 + 1/50} \).- This simplifies to \( \frac{1}{d_i} = \frac{9}{200} \).- Hence, \( d_i = \frac{200}{9} ≈ 22.22 \) cm.This means the image forms 22.22 cm to the right, on the image side of the lens. It's crucial to correctly apply these calculations to predict where images form in lens systems.

Image Height Calculation

In lens systems, knowing an image’s height provides insight into its magnification and orientation. To find an image's height, we use the concept of magnification.The magnification (\( m \)) for any lens is given by: \[ m = -\frac{d_i}{d_o} \]For the first lens:

• With \( d_i = 22.22 \) cm and \( d_o = -50.0 \) cm, the magnification is: \( m = -\frac{22.22}{-50.0} = 0.444 \).
• The original object is 1.20 cm tall, therefore, using the equation: \( h_i = m \times h_o = 0.444 \times 1.20 \).

This results in an image height \( h_i ≈ 0.533 \) cm. This positive height means the image is upright relative to the object.

Lens Magnification

Magnification, represented by \( m \), is a measure of how much larger or smaller the image is compared to the object. It also indicates the image orientation.The formula for magnification is equivalent to the ratio of the image height to the object height, as well as the negative ratio of the image distance to the object distance:\[ m = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \]Key points:

• If \( |m| > 1 \), the image is larger than the object.
• If \( |m| < 1 \), the image is smaller than the object.
• A positive magnification indicates an upright image, while a negative magnification indicates an inverted image.

In our exercise, the magnification for the first lens was 0.444, indicating a smaller, upright image. For the second lens, it was about -0.275, indicating a small, inverted final image.

Focal Length

The focal length is a crucial parameter in lens systems as it determines the converging or diverging power of the lens. It is given in centimeters or meters, depending on the scale of the system. In the exercise: - The first lens had a focal length of 40 cm, indicating shorter convergence, meaning it strongly bends light to a focus point closer to the lens. - The second lens had a focal length of 60 cm, meaning it converges light slightly less intensely, providing a focus farther from the lens. Understanding focal lengths is essential for:

• Designing optical systems, since focal lengths affect the magnification and image positions.
• Predicting how a lens will focus light or how multiple optical components will interact.

This knowledge helps in determining correct placements and combinations of lenses in various practical applications, from microscopes to cameras.