Question

A converging lens with a focal length of 70.0 cm forms an image of a 3.20-cm- tall real object that is to the left of the lens. The image is 4.50 cm tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?

Short answer

The object is 119.50 cm from the lens, and the image is 168.00 cm from the lens on the opposite side. The image is real and inverted.

Step-by-step solution

Understand the problem

We need to find the object distance (\(d_o\)) and image distance (\(d_i\)) using the lens formula and determine whether the image is real or virtual.The given parameters are:- Focal length (\(f\)) = 70 cm- Height of object (\(h_o\)) = 3.20 cm- Height of image (\(h_i\)) = -4.50 cm (inverted image -> negative sign).The lens formula is:\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]The magnification (\(m\)) formula is:\[m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}\]

Use the magnification formula

Calculate the magnification (\(m\)) using the heights of the object and image:\[m = \frac{h_i}{h_o} = \frac{-4.50}{3.20} = -1.40625\]This indicates that the image is inverted and larger by a factor of approximately 1.41.

Solve for image distance (di) with magnification

From the magnification (\(m\)) formula:\[m = -\frac{d_i}{d_o} \Rightarrow d_i = -m \cdot d_o\]Substituting the values:\[d_i = 1.40625 \, d_o\]

Use the lens formula to find object distance (do)

Substitute \(d_i = 1.40625 \, d_o\) into the lens formula:\[\frac{1}{70} = \frac{1}{d_o} + \frac{1}{1.40625 \, d_o}\]Combine terms:\[\frac{1}{70} = \left(\frac{1 + 1.40625}{1.40625 \, d_o}\right)\]Solve for \(d_o\):\[\frac{1}{70} = \frac{2.40625}{1.40625 \, d_o}\]\[d_o = \frac{2.40625 \times 70}{1.40625} \approx 119.50 \text{ cm}\]

Calculate the image distance (di)

Once \(d_o\) is found, calculate \(d_i\) using the relation \(d_i = 1.40625 \, d_o\):\[d_i = 1.40625 \times 119.50 \approx 168.00 \text{ cm}\]

Determine the nature of the image

Since both the object and the image distances are positive, it indicates that a real, inverted image is formed on the opposite side of the lens.

Key concepts

Lens Formula

The lens formula is a fundamental equation used to relate the object distance, the image distance, and the focal length of a lens. For a converging lens, it 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 object distance, and
• \( d_i \) is the image distance.

The formula indicates how the focal length is an inherent property of the lens, while the object and image distances are variables that depend on the positions of the object and the resulting image. Using this equation, we can solve for any one of these variables if the other two are known.
Understanding how to manipulate and apply the lens formula is essential for solving problems involving lenses in physics.

Image Distance

The image distance, denoted as \( d_i \), is the distance from the lens to the image that is formed. It is positive if the image forms on the opposite side of the object in convex (converging) lenses, which results in real images. Conversely, a negative image distance indicates virtual images.

In the given exercise, after using the lens formula and solving the related equations, we find \( d_i \) to be approximately 168 cm. This positive value confirms that the formed image is real and located on the opposite side of the lens from the object.

This concept is crucial as it helps in determining the nature and position of the image formed by lenses, which is valuable for practical applications like in cameras and corrective eyewear.

Magnification

Magnification is the measure of how much larger or smaller the image is compared to the object. It also indicates whether the image is upright or inverted. The magnification \( m \) is calculated by the formula:\[ m = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \]where:

• \( h_i \) is the image height,
• \( h_o \) is the object height,
• \( d_i \) is the image distance, and
• \( d_o \) is the object distance.

In our exercise, the magnification is approximately -1.41, which means the image is about 1.41 times larger than the object and inverted. The negative sign signifies inversion. Understanding magnification helps in predicting how the size and orientation of the image will change relative to the object.
It is a vital parameter in optics, affecting designs in photography, microscopy, and sight correction lenses.

Real and Virtual Images

Images formed by lenses can be either real or virtual. Real images are formed when light rays actually converge at a point, causing the image to be inverted and displaying on a screen. Virtual images, however, occur when light rays only appear to diverge from a point, resulting in an upright image that cannot be projected.

In converging lenses like in this exercise, real images are typically formed when the object is outside of the focal length. For our scenario, since both the object distance \(d_o\) (approximately 119.50 cm) and image distance \(d_i\) (approximately 168 cm) are positive, these indicate a real, inverted image.

• Real Image: Image can be displayed on a screen, inverted, positive \(d_i\).
• Virtual Image: Image cannot be displayed on a screen, upright, negative \(d_i\).

Understanding the nature of these images is fundamental for applications such as optics-based instruments and visual perception systems.