Problem 118 - Fundamentals of Physics

Question

An eraser of height \(1.0 \mathrm{~cm}\) is placed \(10.0 \mathrm{~cm}\) in front of a two-lens system. Lens 1 (nearer the eraser) has focal length \(f_{1}=\) \(-15 \mathrm{~cm},\) lens 2 has \(f_{2}=12 \mathrm{~cm},\) and the lens separation is \(d=12 \mathrm{~cm}\) For the image produced by lens \(2,\) what are (a) the image distance \(i_{2}\) (including sign), (b) the image height, (c) the image type (real or virtual), and (d) the image orientation (inverted relative to the eraser or not inverted)?

Step-by-Step Solution

Answer

(a) 7.2 cm, (b) 1.2 cm, (c) Real, (d) Upright.

Step 1: Calculate Image Position for Lens 1

Use the lens formula \( \frac{1}{f} = \frac{1}{o} + \frac{1}{i} \) for lens 1, where \( f = -15 \, \text{cm} \) and \( o = 10 \, \text{cm} \). Substitute the values to find the image distance \( i_1 \). \[ \frac{1}{-15} = \frac{1}{10} + \frac{1}{i_1} \] Rearrange and solve for \( i_1 \): \[ \frac{1}{i_1} = \frac{1}{-15} - \frac{1}{10} = -\frac{1}{30} \] Therefore, \( i_1 = -30 \, \text{cm} \).

Step 2: Understand the Image from Lens 1

Lens 1 creates a virtual image located 30 cm on the same side as the object. This image acts as a virtual object for lens 2. Because \( i_1 \) is negative, the image is virtual and not inverted.

Step 3: Determine Object Distance for Lens 2

The object distance for lens 2, \( o_2 \), is the lens separation minus the absolute value of \( i_1 \): \( o_2 = d - |i_1| = 12 - 30 = -18 \, \text{cm} \). The negative sign indicates that the object is virtual (on the same side as lens 2).

Step 4: Calculate Image Position for Lens 2

Use the lens formula for lens 2 with \( f = 12 \, \text{cm} \) and \( o_2 = -18 \, \text{cm} \): \[ \frac{1}{12} = \frac{1}{-18} + \frac{1}{i_2} \] Rearrange and solve for \( i_2 \): \[ \frac{1}{i_2} = \frac{1}{12} - \frac{1}{-18} = \frac{5}{36} \] Therefore, \( i_2 = 7.2 \, \text{cm} \).

Step 5: Calculate Image Height for Lens 2

The magnification of lens 1 is \( m_1 = -\frac{i_1}{o_1} = \frac{30}{10} = 3 \). The image height after lens 1 is \( h' = 3 imes 1.0 = 3.0 \, \text{cm} \). For lens 2, the magnification \( m_2 = -\frac{i_2}{o_2} = -\frac{7.2}{-18} = 0.4 \). The final image height \( h'' = m_2 \times 3.0 = 1.2 \, \text{cm} \). Since it's positive, the image is upright.

Step 6: Summary of Image Properties

(a) The image distance \( i_2 = 7.2 \, \text{cm} \) (on the opposite side of lens 2, so positive). (b) The image height is \( 1.2 \, \text{cm} \). (c) The image is real because \( i_2 \) is positive. (d) The image is upright relative to the eraser.

Key Concepts

Image Distance

In a lens system, the image distance is a crucial factor in determining where an image forms relative to a lens. It is denoted by the symbol \(i\) and can be measured from the lens to the location where the image appears. To find the image distance for a given lens, we often use the lens formula:

• \( \frac{1}{f} = \frac{1}{o} + \frac{1}{i} \)

where \(f\) is the focal length of the lens and \(o\) is the object distance from the lens.
The sign of the image distance is important:

• A positive image distance indicates that the image is on the opposite side of the lens as the object and is therefore a real image.
• A negative image distance suggests that the image is on the same side as the object, indicating a virtual image.

A practical example of this can be seen in a two-lens system where lens 1, with a given focal length, creates an image which becomes the virtual object for lens 2. The positioning and calculation of each subsequent image distance are crucial for understanding how a system of lenses will manipulate an incoming light to form an image.

Focal Length

The focal length of a lens is a measure of how strongly it converges or diverges light. It is represented by the symbol \( f \) and is typically measured in centimeters or meters. The focal length is the distance from the lens to the point where parallel light rays meet after being refracted. It's a key element in the lens formula:

• \( \frac{1}{f} = \frac{1}{o} + \frac{1}{i} \)

If the focal length is positive, the lens is converging (or convex), meaning it brings light rays together to focus them, useful in forming real images. Conversely, a negative focal length indicates a diverging (or concave) lens, which spreads light rays apart and typically forms virtual images. In the context of a two-lens system, the different focal lengths of the lenses can significantly impact how light is manipulated, as seen with one lens having a negative focal length and another positive, each contributing both independently and collectively to form an image.

Magnification Calculation

Magnification is the process of enlarging the appearance of an object via optical instruments. It gives the ratio of the image height to the object height and is a dimensionless number. The magnification \( m \) can be calculated using:

• \( m = \frac{h'}{h} = -\frac{i}{o} \)

Here, \( h' \) is the image height and \( h \) is the object height.
Magnification also indicates the orientation of the image:

• A positive magnification means the image is upright relative to the object.
• A negative magnification shows the image is inverted.

In systems with multiple lenses, the total magnification is the product of the magnifications of each lens. This requires calculating the magnification from each lens individually, using the image and object distances. In our example with a multi-lens system, the magnification calculated for lens 1 must be multiplied by the magnification from lens 2 to determine the final image height and orientation.

Virtual and Real Images

Images formed by lenses can be categorized into two types: virtual and real images. The distinction between these two types hinges on the paths of light rays converging:

• Real Images: These occur when light rays actually converge and meet at a point. Such images can be projected onto a screen and are found on the opposite side of the lens from the object. A positive image distance supports the existence of a real image.
• Virtual Images: These are the result of light rays only appearing to diverge from a point. Unlike real images, virtual images cannot be projected on a screen and are located on the same side of the lens as the object. Virtual images are characterized by a negative image distance.

Understanding whether an image is real or virtual is essential because it affects the nature and usability of the image. In practical applications, such as in cameras or the human eye, real images are often desirable. However, virtual images are useful in devices like magnifying glasses or other optical instruments.