Problem 7 - Physics for Scientists and Engineers with Modern Physics

Question

(II) It is desired to magnify reading material by a factor of \(2.5 \times\) when a book is placed 9.0 \(\mathrm{cm}\) behind a lens. (a) Draw a ray diagram and describe the type of image this would be. (b) What type of lens is needed? (c) What is the power of the lens in diopters?

Step-by-Step Solution

Answer

(a) The image is virtual and upright; a convex lens is needed. (b) The lens is a convex lens. (c) The power of the lens is approximately 6.67 diopters.

Step 1: Determine the Lens Type

The magnification of an image is given as negative for real, inverted images and positive for virtual, upright images. Since the magnification desired is positive (2.5), the image is virtual and upright. This indicates we need a converging lens, also known as a convex lens.

Step 2: Calculate the Focal Length of the Lens

We use the lens formula:\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]where \(d_o\) is the object distance and \(d_i\) is the image distance. The magnification \(M\) is given by:\[M = \frac{d_i}{d_o}\]Given \(M = 2.5\) and \(d_o = 9.0\) cm, we find:\[d_i = M \times d_o = 2.5 \times 9.0 \text{ cm} = 22.5 \text{ cm}\]Since the image is virtual for this lens type, \(d_i\) should be negative. Therefore, \(d_i = -22.5\) cm. Substitute \(d_o\) and \(d_i\) into the lens formula:\[\frac{1}{f} = \frac{1}{9.0} - \frac{1}{22.5}\]Calculate \(f\) by solving the equation.

Step 3: Solve for the Focal Length

Continuing from the previous step:\[\frac{1}{f} = \frac{1}{9} - \frac{1}{22.5} = \frac{22.5 - 9}{202.5}\]\[\frac{1}{f} = \frac{13.5}{202.5}\]\[f = \frac{202.5}{13.5} \approx 15.0 \text{ cm} = 0.15 \text{ m}\]We find that the focal length \(f\) is approximately 0.15 m.

Step 4: Calculate the Power of the Lens

The power \(P\) of a lens in diopters is given by the reciprocal of the focal length (in meters):\[P = \frac{1}{f} = \frac{1}{0.15} \approx 6.67 \, \text{D}\]Thus, the power of the lens needed is approximately 6.67 diopters.

Key Concepts

Lens Formula

The lens formula is a crucial aspect of optics, specifically for understanding how lenses form images. It is expressed as \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where:

• \( f \) represents 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 helps in calculating either the focal length, object distance, or image distance when the other two are known. An essential step in using this formula is to sign the distances appropriately. For converging lenses, if an image is virtual, \( d_i \) will be negative because it forms on the same side as the object. This convention is vital for accurate calculations and reinforces understanding of the physical positioning of the image.

Converging Lens

A converging lens, commonly known as a convex lens, has a shape that bends light rays inward. This type of lens is thicker at the center than at the edges. When parallel rays of light pass through a converging lens, they are refracted to meet at a single point known as the focal point. Converging lenses are used in numerous applications, such as in magnifying glasses, eyeglasses for hyperopia, and cameras. They can form both real and virtual images.

• Real images are formed when light rays converge on the opposite side of the lens from where the object is placed. These images are inverted.
• Virtual images appear on the same side as the object. They cannot be projected on a screen and are upright.

In the context of the exercise, the converging lens is responsible for forming a virtual image of the reading material for magnification purposes.

Magnification

Magnification talks about how much larger or smaller the image is when compared to the object. It is calculated using the formula: \( M = \frac{d_i}{d_o} \), which represents the ratio of the image distance \( d_i \) to the object distance \( d_o \). For a virtual and upright image, like in our exercise, the magnification is positive.When the magnification is greater than one, as in this case (2.5 times), it indicates that the image is larger than the object. This property is particularly valuable in reading glasses or other optical tools, facilitating a better view of small or nearby objects. Understanding the sign and value of magnification helps determine the nature of the image formed.

Virtual Image

A virtual image is one of the two primary types of images formed by optical systems. Unlike real images, virtual images cannot be projected onto a screen. They exist in a location where rays of light diverge, creating the illusion of an image. With a converging lens, virtual images:

• Appear on the same side of the lens as the object,
• Are upright relative to the object,
• Result when the object is placed within the focal length of the lens.

The exercise specifies that this type of image is desired to enlarge the text for reading. These images are essential in everyday optical devices like magnifying glasses and certain types of glasses, highlighting the practical application of these theoretical concepts.