Problem 30 - 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 \(\mathrm{m}\) 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) \(d_o \approx 0.0741 \text{ m}\), (b) inverted, (c) \(f \approx 0.0732 \text{ m}\), (d) converging lens.

Step 1: Understand the given information

We are given some distances and magnifications. Let's set them in symbols: Let the object distance be \(d_o\), the image distance be \(d_i = 6.00 \, \text{m} - d_o\) (since the image is 6.00 m away from the slide), and magnification \(m = -80.0\). The negative sign in magnification indicates the image is inverted.

Step 2: Use the magnification formula

We use the magnification formula \( m = \frac{-d_i}{d_o} \). Substituting \(m = -80.0\), we have \(-80.0 = \frac{-(6.00 - d_o)}{d_o}\). Solve for \(d_o\) by cross-multiplying to get \(6.00 - d_o = 80.0 \times d_o\).

Step 3: Solve for object distance \(d_o\)

Simplify the equation from Step 2: \(6.00 = 81d_o\). Divide both sides by 81 to find \(d_o\), giving \(d_o = \frac{6.00}{81} \approx 0.0741 \, \text{m}\).

Step 4: Confirm the image is inverted

Since the magnification (\(m = -80.0\)) is negative, this means the image is inverted.

Step 5: Use lens formula to find focal length

The lens formula is given by \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). With \(d_i = 6.00 - d_o\) and \(d_o \approx 0.0741 \, \text{m}\), substitute these into the formula to find \(f\). Calculate \(\frac{1}{f} = \frac{1}{0.0741 \text{ m}} + \frac{1}{5.926 \text{ m}} \). Simplifying gives \(f \approx 0.0732 \text{ m}\).

Step 6: Determine the type of lens

A diverging lens always forms images that are smaller, thus a magnification greater than 1 (80 in this case) implies it is a converging lens.

Key Concepts

Image Magnification

Image magnification describes how much larger or smaller an image is compared to the actual object. In optics, magnification (\(m\)) is calculated using the formula \(m = \frac{-d_i}{d_o}\), where \(d_i\) is the image distance and \(d_o\) is the object distance. The sign of the magnification tells us the orientation of the image.

• If \(m\) is positive, the image is upright.
• If \(m\) is negative, the image is inverted.

In the given exercise, the magnification is \(-80.0\), meaning the image is 80 times larger than the object and inverted. Understanding magnification is vital to analyze whether you get upside-down images or images of different sizes.

Lens Formula

The lens formula relates the object distance \(d_o\), image distance \(d_i\), and the focal length \(f\) of a lens. It is expressed as \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). This formula is crucial for solving questions involving lenses.

In the exercise, we used this formula to find the focal length after determining the object and image distances. Solving it requires the distances to be known, helping us easily calculate the characteristic, the focal length of the lens. The simplicity of the lens formula makes it a great tool for students, making it easier to solve complex optics problems by breaking down into these distances.

Focal Length

Focal length \(f\) is a measure of how strongly a lens converges or diverges light. It is the distance over which initially collimated rays are brought to a focus.

• In converging lenses, the focal length is positive and is associated with converging light rays towards a focal point.
• In diverging lenses, the focal length is negative and indicates that light rays are spreading outward.

In the context of the homework problem, we found the focal length using the lens equation, resulting in approximately \(0.0732 \text{m}\), signifying a converging lens.

Converging Lens

A converging lens, often called a convex lens, is thicker at the center than at the edges. It has the capability to bring parallel rays of light to a single focal point. These lenses are common in devices like cameras, glasses, and microscopes.

Its defining characteristic is having a positive focal length. Converging lenses can produce magnified images depending on the position of the object relative to the focal point. In the provided problem, the large positive magnification factor indicates the presence of a converging lens, emphasizing its utility in forming large, inverted images from relatively close objects.