Problem 48 - Physics

Question

\(26.3\) at reviews the concepts that play a role in this problem. A converging lens has a focal length of \(88.00 \mathrm{~cm}\). An object \(13.0 \mathrm{~cm}\) tall is located \(155.0\) \(\mathrm{cm}\) in front of this lens. (a) What is the image distance? (b) Is the image real or virtual? (c) What is the image height? Be sure to include the proper algebraic sign.

Step-by-Step Solution

Answer

(a) The image distance is approximately 175.14 cm. (b) The image is real. (c) The image height is approximately -14.68 cm.

Step 1: Understanding the Lens Formula

The lens formula relates the object distance \(d_o\), the image distance \(d_i\), and the focal length \(f\) of a lens. It is given by: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]

Step 2: Substituting Known Values to Find Image Distance

We know that \(f = 88.00 \text{ cm}\) and the object distance \(d_o = 155.0 \text{ cm}\). We need to find \(d_i\). Substitute these values into the lens formula: \[ \frac{1}{88.00} = \frac{1}{155.0} + \frac{1}{d_i} \] After rearranging the formula, solve for \(\frac{1}{d_i}\): \[ \frac{1}{d_i} = \frac{1}{88.00} - \frac{1}{155.0} = \frac{0.01136}{\text{approximate value}} \] This results in \(d_i \approx 175.14 \text{ cm}\).

Step 3: Determining the Nature of the Image

An image is real if the image distance \(d_i\) is positive, and virtual if \(d_i\) is negative. Since \(d_i = 175.14 \text{ cm}\) is positive, the image is real.

Step 4: Finding the Image Height

Use the magnification formula, which relates the object height \(h_o\), the image height \(h_i\), the object distance \(d_o\), and the image distance \(d_i\). The formula is given by: \[ m = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \] Given \(h_o = 13.0 \text{ cm}\), \(d_i = 175.14 \text{ cm}\), and \(d_o = 155.0 \text{ cm}\), substitute these into the magnification formula to solve for \(h_i\): \[ \frac{h_i}{13.0} = -\frac{175.14}{155.0} \] Hence, \(h_i = -\frac{175.14}{155.0} \cdot 13.0 \approx -14.68 \text{ cm}\). The negative sign indicates that the image is inverted.

Key Concepts

Lens Formula

The lens formula is crucial for understanding how images are formed by lenses. It is a mathematical equation that relates the focal length of the lens, the distance from the object to the lens, and the distance from the image to the lens. The formula is written as: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Where:

• \( f \) is the focal length of the lens.
• \( d_o \) is the distance from the object to the lens.
• \( d_i \) is the distance from the image to the lens.

This formula is derived from the principle of optical lenses, where light converges or diverges to form images. It is essential to identify the focal length correctly, which is positive for converging lenses and negative for diverging lenses.
Understanding this formula helps in calculating other significant properties of images, such as their size and orientation.

Image Distance

Image distance is an important concept in understanding how lenses work. It refers to how far the formed image is from the lens. In the context of the lens formula, the image distance \( d_i \) can be calculated if the focal length \( f \) and object distance \( d_o \) are known.
In solving for \( d_i \), we frequently rearrange the lens formula:\[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \]By solving this equation, the value of \( d_i \) can be found. In our example, with a focal length of \( 88.00 \) cm and an object distance of \( 155.0 \) cm, \( d_i \) came out to approximately \( 175.14 \) cm.
Importantly, the sign of \( d_i \) reveals more than just where the image is located. A positive \( d_i \) indicates a real image, which means it forms on the opposite side of the lens compared to where the object is placed. Conversely, when \( d_i \) is negative, it represents a virtual image.

Magnification

Magnification describes how much larger or smaller the image is compared to the actual object. This is an essential factor when analyzing optical lenses because it allows us to predict the size and orientation of the image. The magnification \( m \) is given by:\[ m = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \]Where:

• \( h_i \) is the height of the image.
• \( h_o \) is the height of the object.
• \( d_i \) and \( d_o \) are image and object distances, respectively.

For our lens example:- The object was \( 13.0 \) cm tall, creating an image with a height calculated to be approximately \( -14.68 \) cm using the formula. - The negative magnification value indicates that the image is inverted.
Magnification can significantly affect what we perceive through optical lenses, whether in microscopes, glasses, or cameras. Understanding this allows us to adjust and modify lens systems for desired outcomes.