Question

Two converging lenses are placed \(88.0 \mathrm{cm}\) apart. An object is placed \(1.100 \mathrm{m}\) to the left of the first lens. which has a focal length of \(25.0 \mathrm{cm} .\) The final image is located \(15.0 \mathrm{cm}\) to the right of the second lens. (a) What is the focal length of the second lens? (b) What is the total magnification?

Short answer

The focal length of the second lens is approximately 11.95 cm, and the total magnification is approximately 0.089.

Step-by-step solution

Determine the image formed by the first lens

The object distance for the first lens, \(d_{o1}\), is \(1.100\ \text{m} = 110.0\ \text{cm}\). Using the lens formula \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where the focal length \(f_1\) is \(25.0\ \text{cm}\), find the image distance, \(d_{i1}\), using:\[\frac{1}{25} = \frac{1}{110} + \frac{1}{d_{i1}}\]Solve for \(d_{i1}\):\[d_{i1} = \frac{1}{\frac{1}{25} - \frac{1}{110}} = 34.72\ \text{cm} \text{ (approx)}\].

Determine the object distance for the second lens

The image formed by the first lens serves as the object for the second lens. The distance between the lenses is \(88.0\ \text{cm}\). Thus, the object distance for the second lens, \(d_{o2}\), is:\[d_{o2} = 88.0 - 34.72 = 53.28\ \text{cm}\].

Use the final image position to find the second lens's focal length

The final image distance from the second lens, \(d_{i2}\), is given as \(15.0\ \text{cm}\) to the right, so it is \(d_{i2} = 15.0\ \text{cm}\). Using the lens formula again for the second lens,\[\frac{1}{f_2} = \frac{1}{d_{o2}} + \frac{1}{d_{i2}}\]Substitute the known values,\[\frac{1}{f_2} = \frac{1}{53.28} + \frac{1}{15}\]Solve for \(f_2\):\[f_2 = \frac{1}{\frac{1}{53.28} + \frac{1}{15}} = 11.95\ \text{cm} \text{ (approx)}\].

Calculate the magnification for each lens and the total magnification

The magnification for the first lens is given by:\[m_1 = -\frac{d_{i1}}{d_{o1}} = -\frac{34.72}{110} = -0.316 (\text{approx})\]The magnification for the second lens is:\[m_2 = -\frac{d_{i2}}{d_{o2}} = -\frac{15}{53.28} = -0.281\]The total magnification, \(M\), is the product of \(m_1\) and \(m_2\):\[M = m_1 \times m_2 = (-0.316) \times (-0.281) = 0.089(\text{approx})\].

Key concepts

Focal Length Calculation

Understanding the focal length of a lens is pivotal in optics. This value determines how strongly the lens converges or diverges light. The process to find the focal length involves employing the lens formula:

• \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance.

In our exercise with two converging lenses, we already know the details for the first lens: object's distance \( d_{o1} \) is 110 cm and the image distance \( d_{i1} \) was calculated as approximately 34.72 cm. The lens's focal length \( f_1 \) is provided as 25 cm. For the second lens, which is placed 88 cm apart from the first one, the focal length needs to be calculated using the known object distance \( d_{o2} \) = 53.28 cm (as derived from the system) and the final image distance \( d_{i2} \) = 15 cm.Calculating the focal length \( f_2 \) involves:

• Plug-in values into the lens formula: \( \frac{1}{f_2} = \frac{1}{53.28} + \frac{1}{15} \)
• Solve for \( f_2 \): \( f_2 = \left(\frac{1}{53.28} + \frac{1}{15}\right)^{-1} = 11.95 \ \text{cm (approx)} \)

This gives us the result that the second lens has a focal length of approximately 11.95 cm.

Image Distance Determination

Image distance is crucial in understanding where an image forms relative to a lens. To determine this, we once again rely on the lens formula. Let's walk through the steps for both lenses in our exercise:- **First Lens**: - Given: Object distance \( d_{o1} = 110 \ \text{cm}\) and focal length \( f_1 = 25 \ \text{cm}\). - Using the lens formula, solve for image distance \( d_{i1} \): \[ \frac{1}{25} = \frac{1}{110} + \frac{1}{d_{i1}} \ d_{i1} = \left( \frac{1}{25} - \frac{1}{110} \right)^{-1} \approx 34.72 \ \text{cm} \]- **Second Lens**: - The image from the first lens becomes the object for the second. We determined the new object distance \( d_{o2} = 53.28 \ \text{cm} \) through the relation \( 88 - 34.72 \). - Given the final image distance: \( d_{i2} = 15 \ \text{cm} \). - Substitute and solve to verify consistency: \[ \frac{1}{f_2} = \frac{1}{53.28} + \frac{1}{15} \ \] Following this process confirms the locations where the images form. It's essential for predicting how lenses affect light paths and create images.

Magnification Calculation

Magnification describes how much larger or smaller an image is compared to the object itself. It provides insight into the type of image produced—whether it is inverted or upright. This concept is mathematically expressed as:

• \( m = -\frac{d_i}{d_o} \)

In a system of two lenses, total magnification is the product of individual magnifications from both lenses.For the first lens:- Given: \( d_{i1} = 34.72 \ \text{cm} \) and \( d_{o1} = 110 \ \text{cm} \).- Calculate magnification \( m_1 \): \[ m_1 = -\frac{34.72}{110} \approx -0.316 \]For the second lens:- Given: \( d_{i2} = 15 \ \text{cm} \) and \( d_{o2} = 53.28 \ \text{cm} \).- Calculate magnification \( m_2 \): \[ m_2 = -\frac{15}{53.28} \approx -0.281 \]Finally, determine the total magnification of the system:\[M = m_1 \times m_2 = (-0.316) \times (-0.281) = 0.089\]This calculation tells us that the image is slightly reduced in size compared to the original object. Understanding this allows us to predict how systems of multiple lenses impact visual output in applications such as telescopes and cameras.