Problem 32 - University Physics with Modern Physics

Question

A photographic slide is to the left of a lens. The lens projects an image of the slide onto a wall 6.00 m to the right of the slide. The image is 80.0 times the size of the slide. (a) How far is the slide from the lens? (b) Is the image erect or inverted? (c) What is the focal length of the lens? (d) Is the lens converging or diverging?

Step-by-Step Solution

Answer

(a) 0.074 m, (b) inverted, (c) 0.073 m, (d) converging lens.

Step 1: Understand the Problem

We are given that the image is 6.00 m to the right of the slide and is 80.0 times larger than the slide itself. We need to find the distance of the slide from the lens, determine the nature (erect or inverted) and type (converging or diverging) of the lens, and calculate its focal length.

Step 2: Use Magnification Formula

The magnification of the image, denoted as \( M \), is given by \( M = \frac{h'}{h} = \frac{-d_i}{d_o} \), where \( h' \) and \( h \) are the image and object heights, \( d_i \) is the image distance, and \( d_o \) is the object distance. We know \( M = -80 \) (since the image is magnified and typically inverted for real images), and \( d_i = 6 - d_o \). Solve \( -80 = \frac{-d_i}{d_o} \) to find \( d_o \).

Step 3: Solve for Slide Distance (a)

Using \( -80 = \frac{-(6 - d_o)}{d_o} \), we rearrange to get \( 80d_o = 6 - d_o \). Solve for \( d_o \) by \( 80d_o + d_o = 6 \), which simplifies to \( 81d_o = 6 \). So, \( d_o = \frac{6}{81} = \frac{2}{27} \approx 0.074 \text{ m} \).

Step 4: Determine Image Characteristics (b)

Since the magnification is negative, the image is inverted (as is typical for real images formed by lenses).

Step 5: Use Lens Formula for Focal Length (c)

The lens formula is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). We know \( d_o = 0.074 \text{ m} \) and \( d_i = 6 - 0.074 = 5.926 \text{ m} \). Calculate \( \frac{1}{f} = \frac{1}{0.074} + \frac{1}{5.926} \). This gives \( \frac{1}{f} = 13.51 + 0.169 \approx 13.679 \). Thus, \( f \approx \frac{1}{13.679} \approx 0.073 \text{ m} \).

Step 6: Determine Lens Type (d)

Since the focal length is positive, the lens is converging (convex).

Key Concepts

Lens Formula

The lens formula is a fundamental equation in geometrical optics that connects the focal length of a lens to the distances of the object and the image from the lens. The formula is expressed 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, and
• \( d_i \) is the distance from the image to the lens.

This formula is applicable to both converging (convex) and diverging (concave) lenses, but the sign conventions differ. For converging lenses, the focal length is positive, while for diverging lenses it is negative. This straightforward relation helps in determining unknown distances when the focal length is given, or vice versa.
In our exercise, we used this formula to calculate the focal length based on known distances, ensuring the negative image magnification is consistent with a real, inverted image.

Image Magnification

Image magnification allows us to understand how much larger or smaller an image is compared to the original object. In optics, the magnification \( M \) can be expressed as:
\[ M = \frac{h'}{h} = \frac{-d_i}{d_o} \]
where:

• \( h' \) and \( h \) are the heights of the image and object, respectively,
• \( d_i \) is the image distance,
• \( d_o \) is the object distance.

The negative sign indicates the inversion of the image. This means that an upside-down (or inverted) image forms when the magnification is negative. In our scenario, the image was magnified by 80 times, suggesting a significant enlargement and inversion due to a converging lens. Using this concept, we solved for the object distance knowing the image magnification and distance.

Converging Lens

A converging lens, often called a convex lens, has the ability to focus parallel light rays to a single point known as the focal point. This characteristic is crucial in forming sharp images of objects. Converging lenses are typically thicker at the center than at the edges, creating their unique shape.
These lenses are used in a wide range of applications, such as in cameras, eyeglasses for correcting farsightedness, and projectors.
In the context of our exercise:

• The positive focal length calculated indicates a converging lens.
• The lens inverts and magnifies the image, projecting a clear image onto a distant screen.

By understanding the properties of a converging lens, we can predict how light behaves as it transits through the lens, giving us plenty of useful applications in optics.

Inverted Image

An inverted image is one that is flipped upside down relative to the original object. This flipping is a common occurrence in optical systems using lenses and mirrors. In our problem:

• The image formed by the lens is inverted, as shown by the negative magnification factor.
• This inversion is typical when real images are produced by converging lenses given the nature of light ray convergence.

The inversion occurs because the top of the object is imaged at the bottom and vice versa due to the crossing of light rays at the focal point.
This understanding of image characteristics helps differentiate between different types of lenses and their uses. Inverted images, although upside down, can be captured by screens or photographic sensors, allowing the processing of visual information both in scientific and everyday visual applications like photography.