Problem 120 - Fundamentals of Physics

Question

A coin is placed \(20 \mathrm{~cm}\) in front of a two-lens system. Lens 1 (nearer the coin) has focal length \(f_{1}=+10 \mathrm{~cm}\), lens 2 has \(f_{2}=\) \(+12.5 \mathrm{~cm}\), and the lens separation is \(d=30 \mathrm{~cm} .\) For the image produced by lens 2, what are (a) the image distance \(i_{2}\) (including sign), (b) the overall lateral magnification, (c) the image type (real or virtual), and (d) the image orientation (inverted relative to the coin or not inverted)?

Step-by-Step Solution

Answer

a) \(i_2 \approx 5.56\text{ cm}\), b) \(m = -0.556\), c) Real, d) Inverted

Step 1: Calculate the Image Distance for Lens 1

Lens 1's focal length is given by \( f_1 = +10 \text{ cm} \), and the object distance \( o_1 \) is \( 20 \text{ cm} \). Using the lens formula \( \frac{1}{f} = \frac{1}{o} + \frac{1}{i} \), we can find the image distance for lens 1 \( i_1 \): \[\frac{1}{10} = \frac{1}{20} + \frac{1}{i_1}\]Solving for \( i_1 \):\[\frac{1}{i_1} = \frac{1}{10} - \frac{1}{20} = \frac{1}{20}\]\[ i_1 = 20 \text{ cm} \] (Positive value implies image is real and on the opposite side from the object).

Step 2: Adjust Image Distance for Lens 2

The image by lens 1 now becomes the object for lens 2. The distance from this image to lens 2 is \( d - i_1 = 30 \text{ cm} - 20 \text{ cm} = 10 \text{ cm} \). Because this image is on the opposite side for lens 2, the object distance \( o_2 \) is negative: \( o_2 = -10 \text{ cm} \).

Step 3: Calculate the Image Distance for Lens 2

Using lens 2's focal length \( f_2 = +12.5 \text{ cm} \), and object distance for lens 2 \( o_2 = -10 \text{ cm} \), apply the lens formula:\[\frac{1}{12.5} = \frac{1}{-10} + \frac{1}{i_2}\]Solving: \[\frac{1}{i_2} = \frac{1}{12.5} + \frac{1}{10} = \frac{2.5 + 2}{25} = \frac{4.5}{25}\]\[ i_2 = \frac{25}{4.5} \approx 5.56 \text{ cm} \] (Positive value indicates the image is real).

Step 4: Calculate the Overall Lateral Magnification

The magnification produced by each lens is given by \( m_1 = -\frac{i_1}{o_1} \) and \( m_2 = -\frac{i_2}{o_2} \). First, calculate \( m_1 \):\[m_1 = -\frac{20}{20} = -1\]Then calculate \( m_2 \):\[m_2 = -\frac{5.56}{-10} = 0.556\]The overall magnification is \( m = m_1 \times m_2 = -1 \times 0.556 = -0.556 \).

Step 5: Determine the Image Type

The image type is determined by the sign of \( i_2 \). Since \( i_2 \approx 5.56 \text{ cm} \) is positive, the image produced by lens 2 is real.

Step 6: Determine the Image Orientation

The orientation is determined by the sign of the overall magnification. Since \( m = -0.556 \) is negative, the image is inverted relative to the original object.

Key Concepts

Lens Formula

The lens formula is fundamental in geometric optics, describing the relationship between the focal length of a lens, the object distance, and the image distance. It is given by the equation:\[ \frac{1}{f} = \frac{1}{o} + \frac{1}{i} \]where

• \( f \) is the focal length of the lens,
• \( o \) is the object distance (distance from the object to the lens),
• \( i \) is the image distance (distance from the image to the lens).

This formula allows us to predict where an image will form and whether it will be real or virtual, which is crucial when working with complex optical systems like a two-lens setup. The signs for these values are determined based on the conventions of the optical setup (real and virtual sides of the lens).

Image Distance Calculation

In problems involving lenses, calculating the image distance is a necessary step, often dictated by the lens formula. For a two-lens system, it involves a sequence of calculations.
The initial task is finding the image distance for Lens 1 using its focal length and the object's position in front of it. With the lens formula:\[ \frac{1}{f_1} = \frac{1}{o_1} + \frac{1}{i_1} \]For Lens 1 with a focal length of \( f_1 = +10 \text{ cm} \) and object distance \( o_1 = 20 \text{ cm} \), calculate as follows:

• Solve for \( i_1 \) to determine image location.

Once \( i_1 \) is determined, it becomes the object for Lens 2, taking into account the separation between the lenses. The formula requires adjusting the object distance for Lens 2:\[ \frac{1}{f_2} = \frac{1}{o_2} + \frac{1}{i_2} \]By accurately performing these calculations, we can deduce where the image will form relative to the second lens.

Lateral Magnification

Lateral magnification gives us insight into the size change of an image relative to the object it originates from. It is calculated using the image and object distances:\[ m = -\frac{i}{o} \]For a multi-lens system, each lens contributes to the total magnification. We calculate the magnification for each lens separately:

• For Lens 1: \( m_1 = -\frac{i_1}{o_1} \)
• For Lens 2: \( m_2 = -\frac{i_2}{o_2} \)

The final step is to multiply these individual magnifications to obtain the total magnification:\[ m = m_1 \times m_2 \]The magnitude of this value tells us how much larger or smaller the image is compared to the original object, while the sign indicates if the image is inverted.

Real and Virtual Images

Understanding the nature of real and virtual images is key in optics. Real images occur when light converges to form an image on the real side of a lens, and they can be projected onto a screen. They have a positive image distance. In contrast, virtual images appear when light rays appear to diverge from a point. These images form on the same side as the object with a negative image distance and cannot be projected.In the context of the exercise, after calculating the image distance \( i_2 \) for the second lens, the positive value indicates the image is real. Such information helps us predict whether an image in a lens system like ours will be inverted and can be seen without a screen.