Problem 27 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A lighted candle is placed 36 \(\mathrm{cm}\) in front of a converging lens of focal length \(f_{1}=13 \mathrm{cm},\) which in turn is 56 \(\mathrm{cm}\) in front of another converging lens of focal length \(f_{2}=16 \mathrm{cm}\) (see Fig. \(47 ) .(a)\) Draw a ray diagram and estimate the location and the relative size of the final image. (b) Calculate the position and relative size of the final image.

Step-by-Step Solution

Answer

The final image forms 24.08 cm past Lens 2, reduced to 0.39 times the object size (upright).

Step 1: Understand the Lens System

The system consists of two lenses: Lens 1 with focal length \( f_1 = 13 \text{ cm} \) and Lens 2 with focal length \( f_2 = 16 \text{ cm} \). The object (candle) is placed 36 cm in front of Lens 1, and the lenses are 56 cm apart. We will calculate where the image forms after passing through both lenses.

Step 2: Use Lens Formula for First Lens

Apply the lens formula \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) to the first lens. Here, \( d_o = 36 \text{ cm} \) and \( f_1 = 13 \text{ cm} \). Calculate \( d_{i1} \) (image distance from the first lens): \[\frac{1}{13} = \frac{1}{36} + \frac{1}{d_{i1}} \\frac{1}{d_{i1}} = \frac{1}{13} - \frac{1}{36} \d_{i1} \approx 20.47 \text{ cm}\]

Step 3: Determine the Object Distance for Second Lens

The image formed by the first lens acts as the virtual object for the second lens. Since the lenses are 56 cm apart, the object distance for the second lens is calculated as \( d_{o2} = 56 - d_{i1} \): \[d_{o2} = 56 - 20.47 = 35.53 \text{ cm}\]

Step 4: Use Lens Formula for Second Lens

Apply the lens formula to the second lens with \( d_{o2} = 35.53 \text{ cm} \) and \( f_2 = 16 \text{ cm} \): \[\frac{1}{16} = \frac{1}{35.53} + \frac{1}{d_{i2}} \\frac{1}{d_{i2}} = \frac{1}{16} - \frac{1}{35.53} \d_{i2} \approx 24.08 \text{ cm}\]Thus, the final image forms approximately 24.08 cm from the second lens.

Step 5: Calculate the Magnification

The overall magnification \(M\) is the product of the magnifications from each lens. Calculate the magnification for each lens: For Lens 1: \( M_1 = \frac{-d_{i1}}{d_o} = \frac{-20.47}{36} \approx -0.57 \)For Lens 2: \( M_2 = \frac{-d_{i2}}{d_{o2}} = \frac{-24.08}{35.53} \approx -0.68 \)The overall magnification is \( M = M_1 \times M_2 \approx -0.57 \times -0.68 \approx 0.39 \).This means the final image is 0.39 times the size of the object (reduced and upright).

Step 6: Draw the Ray Diagram

Draw a ray diagram illustrating the system: 1. From the object, draw rays through Lens 1: one parallel to the principal axis through the focal point on the opposite side, and another through the center of the lens. Mark the image formed by Lens 1. 2. Use this image as the object for Lens 2: draw similar rays through Lens 2. Determine the final image's position and relative size approximately based on the diagram. The ray diagram should illustrate light bending at each lens, producing an image after each lens.

Key Concepts

Converging Lens

A converging lens, commonly known as a convex lens, is an integral part of optical systems used to focus light. These lenses are thicker at the center than at the edges. They work by collecting light rays that are initially coming parallel to each other and converge them to a single point, known as the focal point. This ability to focus light makes them especially useful in applications like magnifying glasses and corrective lenses.

Key characteristics of a converging lens include:

• Focal length ( ): The distance from the lens to the focal point. It is positive for converging lenses and indicates the strength of their ability to bend light.


• Focal point: The spot where parallel light rays converge after passing through the lens.


• Principal axis: The straight line passing through the center and the focal points of the lens.

In the given exercise, two converging lenses are utilized. Each has a specified focal length—13 cm for the first lens and 16 cm for the second lens—which affects how they process the light from the object.

Lens Formula

The lens formula is essential for understanding how lenses form images. It relates the object distance (), image distance (), and the focal length () of a lens through the equation:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]

Here's what each term means:

• : The distance between the object and the lens.


• : The distance between the image and the lens.


• : The focal length, a known value for converging lenses which helps determine where the image will form.

This formula helps calculate the image distance (), which tells you where the image will form relative to the lens. In the exercise, this formula is applied twice, once for each lens, to ascertain the final position of the image. Calculating the image distance for the first lens provides a virtual object location for the second lens, continuing the process through the lens system.

Ray Diagram

A ray diagram is a graphical method used to trace the path of light through lenses, illustrating how images form. It's a powerful tool for visualizing the process of image formation by a lens or a series of lenses.

To draw a ray diagram for a converging lens system, like the one in the problem, follow these steps:

• Start by identifying the principal axis, focal point, and optical center of the lens.


• For the first lens, draw a ray emerging parallel to the principal axis that will pass through the focal point on the lens's opposite side.


• Draw a second ray that goes straight through the center of the lens without bending, marking the first image location where these rays meet.

Repeat this process using the image formed by the first lens as the new "object" for the second lens. This method allows you to determine both the position and the size of the final image. Ray diagrams provide intuitive understanding by showcasing light pathways, making abstract calculations more concrete and less confusing.