Question

An unknown lens forms an image of an object that is \(24 \mathrm{~cm}\) away from the lens, inverted, and a factor of 4 larger in size than the object. Where is the object located? a) \(6 \mathrm{~cm}\) from the lens on the same side of the lens b) \(6 \mathrm{~cm}\) from the lens on the other side of the lens c) \(96 \mathrm{~cm}\) from the lens on the same side of the lens d) \(96 \mathrm{~cm}\) from the lens on the other side of the lens e) No object could have formed this image.

Short answer

Answer: The object is located 6 cm from the lens on the same side of the lens.

Step-by-step solution

Write down the lens formula and magnification formula for lenses

The lens formula is given by: $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ Where \(f\) is the focal length of the lens, \(v\) is the image distance, and \(u\) is the object distance. The magnification formula is given by: $$m = \frac{v}{u}$$

Substitute the given information into the magnification formula

We are given the magnification \(m = 4\) and the image distance \(v = 24 \mathrm{~cm}\). Substitute these values into the formula: $$4 = \frac{24 \mathrm{~cm}}{u}$$

Solve for the object distance, u

Solve for \(u\) in the equation: $$u = \frac{24 \mathrm{~cm}}{4}$$ $$u = 6 \mathrm{~cm}$$ So, the object is \(6 \mathrm{~cm}\) away from the lens.

Determine the side of the lens where the object is placed

Since the image is inverted and larger, the lens has to be a converging lens, and the object must be placed in front of the lens. Therefore, the object is placed on the same side of the lens as the image. The correct answer is: (a) \(6 \mathrm{~cm}\) from the lens on the same side of the lens.

Key concepts

Image Magnification

In optics, the term image magnification refers to the ratio of the size of the image produced by a lens to the size of the object that was captured. Mathematically, magnification (\(m\)) is expressed as the ratio of the image distance (\(v\)) to the object distance (\(u\)), represented by the equation: \[\begin{equation}m = \frac{v}{u}\end{equation}\]In the context of the exercise provided, where an image is formed that is four times larger than the object, we use a magnification value of 4. This implies that the ratio of the image's distance from the lens to the object's distance from the lens is 4 to 1. Therefore, if the image is located 24 cm from the lens, the object's distance can be determined using the formula
\[\begin{equation}4 = \frac{24 \text{ cm}}{u}\end{equation}\]
Solving for the object distance (\(u\)), we find that it is 6 cm. Understanding magnification is critical for solving problems related to image formation in lenses, as it directly influences the perceived size of the image and helps to deduce the position of the object relative to the lens.

Converging Lens

A converging lens is a type of lens that bends light rays such that they meet at a point after passing through the lens. This point is known as the focal point, and the distance from the lens to the focal point is the focal length (\(f\)). Converging lenses are thicker at the center and thinner at the edges, and they are commonly used in a variety of optical devices such as cameras, telescopes, and eyeglasses.

In our exercise, the key indicator that we are dealing with a converging lens is the nature of the image formed: it is inverted and magnified. These characteristics imply that the object is located at a distance greater than the focal length from the lens but within twice the focal length (between the focal point and the center of curvature). For a converging lens, any object placed within this range will produce a real, inverted, and magnified image on the same side of the lens as the object. This behavior is distinct from a diverging lens, which would produce a virtual, upright, and reduced image.

Optics in Physics

The study of optics in physics encompasses the behavior and properties of light, including its interactions with matter and the construction of instruments that use or detect it. Optics is divided into various subfields such as geometric optics, which deals with light propagation in terms of rays, and wave optics, which takes into account the wave nature of light.

Geometric optics, in particular, is fundamental when working with lenses and mirrors, where the laws of reflection and refraction are applied to predict how light will travel. It's essential for students to grasp these principles, as they are the groundwork for understanding phenomena such as image formation by lenses, characterized by the lens formula \[\begin{equation}\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\end{equation}\]

1. Reflection: the bouncing of light when it hits a surface.
2. Refraction: the bending of light as it passes from one medium to another with a different density.

By combining theoretical knowledge with practical exercises like the one we've discussed, students can better comprehend and apply these concepts, allowing them to effectively analyze and solve complex problems in optics.