Problem 31 - College Physics

Question

. The focal length of a simple magnifier is 8.00 \(\mathrm{cm} .\) Assume the magnifier to be a thin lens placed very close to the eye. (a) How far in front of the magnifier should an object be placed if the image is formed at the observer's near point, 25.0 \(\mathrm{cm}\) in front of her eye? (b) If the object is 1.00 \(\mathrm{mm}\) high, what is the height of its image formed by the magnifier?

Step-by-Step Solution

Answer

(a) 13.3 cm; (b) -1.88 mm (inverted image).

Step 1: Understand the Given Information

We have a magnifier with a focal length \(f = 8.00\, \mathrm{cm}\). The image is formed at the observer's near point \(d_i = 25.0\, \mathrm{cm}\) in front of the eye. We need to find the object distance \(d_o\).

Step 2: Use the Lens Formula

The lens formula relates object distance \(d_o\), image distance \(d_i\), and focal length \(f\): \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]. Substitute \(f = 8.00\, \mathrm{cm}\) and \(d_i = 25.0\, \mathrm{cm}\) into the formula.

Step 3: Solve for Object Distance

Rearrange the lens formula to solve for \(d_o\): \[\frac{1}{d_o} = \frac{1}{f} - \frac{1}{d_i} = \frac{1}{8.00} - \frac{1}{25.0}\]. Calculate \(d_o\) using the values: \(\frac{1}{d_o} = \frac{1}{8.00} - \frac{1}{25.0}\), so \(d_o = \frac{1}{0.075} \approx 13.3\, \mathrm{cm}\).

Step 4: Calculate Image Height Using Magnification

The magnification \(m\) is given by \(m = -\frac{d_i}{d_o}\). Calculate the magnification using \(d_i = 25.0\, \mathrm{cm}\) and \(d_o = 13.3\, \mathrm{cm}\): \(m = -\frac{25.0}{13.3}\).

Step 5: Find Image Height

The height of the image \(h_i\) can be found from the magnification \(m = \frac{h_i}{h_o}\), where \(h_o = 1.00\, \mathrm{mm}\). Solve for \(h_i\): \(h_i = m \cdot h_o = -\frac{25.0}{13.3} \cdot 1.00 \approx -1.88\, \mathrm{mm}\). The negative sign indicates the image is inverted.

Key Concepts

Focal Length

The focal length of a lens is a critical aspect that determines how the lens converges or diverges light. It's represented by the symbol \( f \) and is usually measured in centimeters or meters, depending on the context. For a convex lens, like a magnifying glass, the focal length is positive, indicating converging behavior. This means that when parallel rays of light pass through the lens, they converge at a point known as the focus. The focal length is the distance from the center of the lens to this focal point.

In our exercise, the focal length is given as \( 8.00 \, \text{cm} \). This means that if you place an object infinitely far away (like the sun), the rays will converge \( 8 \text{ cm} \) away from the lens. Understanding the focal length helps in applications such as focusing cameras, glasses, and of course, magnifiers.

Magnification

Magnification is a measure of how much larger or smaller an image is compared to the object. It is calculated as the ratio of the image height to the object height and is represented by the symbol \( m \). A magnification greater than one indicates an enlarged image, while less than one means the image is reduced.

The magnification is also related to the object and image distances, noted as \( m = -\frac{d_i}{d_o} \), where \( d_i \) is the image distance and \( d_o \) is the object distance. The negative sign indicates the image is inverted with respect to the object.

In this exercise, the image is formed at the observer's near point which is \( 25.0 \, \text{cm} \) in front of her eye, and the object is placed \( 13.3 \, \text{cm} \) from the lens. Using these distances, we find the magnification to be approximately \( -1.88 \), signifying the image is both larger and inverted compared to the object.

Object Distance

Object distance refers to how far the object is from the lens, denoted as \( d_o \). It is a crucial parameter when using lenses as it influences both the size and orientation of the produced image. To accurately locate where the object should be placed, one often uses the lens formula:

\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]

In our scenario, after inputting the known values, the object should be positioned \( 13.3 \, \text{cm} \) in front of the magnifier so that the image appears exactly at the observer's near point. Calculations with the lens formula allow us to predict and control how lenses behave in practical situations like reading glasses and magnifying glasses.

Image Formation

Image formation by lenses involves understanding how light travels and creates an image through refraction. When light rays pass through a lens, they bend and converge or diverge to form an image. The nature of the image – whether it appears real or virtual, upright or inverted, enlarged or reduced – depends on several factors, including the lens type, object position, and focal length.

With our setup, we have a thin lens (the magnifier) and the image is formed at \( 25.0 \, \text{cm} \) from the observer's eye, which is a typical near point for comfortable viewing. By using the lens formula and magnification principles, this exercise shows that the resulting image is inverted and larger than the object. This means the light rays indeed converge, producing an enlarged, upside-down virtual image that appears closer and bigger to the observer, enhancing the details visible through the magnifier.