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At what distance \(d_o\) should an object be placed in front of a converging lens such that the image is at the same distance on the other side of the lens? (A) \(d_o=f\) (B) \(d_o=\infty\) (C) \(d_o=f / 2\) (D) \(d_o=2 f\) (E) It cannot be done

Short Answer

Expert verified
Answer: The object distance in this case is \(d_o=2f\).

Step by step solution

01

Write down the lens formula

We'll start by writing down the lens formula: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\)
02

Set \(d_o = d_i\)

Since we know that the object distance and image distance are equal, we can set \(d_o = d_i\). We will now have the equation: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_o}\)
03

Simplify and solve for \(d_o\)

Combining the terms on the right side of the equation, we get: \(\frac{1}{f} = \frac{2}{d_o}\). Now, we can solve for \(d_o\) by multiplying both sides of the equation with \(d_o\) and then dividing by \(\frac{1}{f}\), which gives: \(d_o = \frac{2}{\frac{1}{f}}\)
04

Calculate the value of \(d_o\)

Finally, we can rewrite the equation for \(d_o\) in terms of \(f\): \(d_o = 2f\)
05

Compare the result with the given options

The expression we found for \(d_o\) is \(2f\), which matches option (D). Therefore, the correct answer is (D) \(d_o=2f\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Converging Lens
Understanding the properties of a converging lens is crucial for solving various optics problems. A converging lens, also known as a convex lens, bends light rays in such a way that they meet or converge at a point. When parallel rays of light, such as those coming from the sun, pass through a converging lens, they refract inwards and meet at the focal point on the other side. The focal point, thus, is the exact spot where the light rays would theoretically converge.

This behavior of converging lenses makes them widely used for focusing light in devices like cameras, projectors, and eyeglasses. In physics, working with the notion that light behaves predictably through these lenses allows us to create mathematical models, predict how the light will behave, and construct systems like microscopes and telescopes with precise magnification capabilities.

One key characteristic of a converging lens is its ability to produce both real and virtual images, depending on the position of the object relative to the focal length (which we’ll discuss in a later section). It’s important to remember that real images are formed on the side of the lens opposite the object and can be projected onto a screen, while virtual images appear on the same side as the object and cannot be projected.
Object Distance
In lens formulas, the object distance, denoted by \(d_o\), is the distance between the object and the lens. It is an integral variable in determining the behavior of light rays as they pass through lenses. An object's position relative to a lens's focal length can greatly influence the size, orientation, and type of image formed.

The object distance, along with the image distance and focal length, is used in the lens formula to calculate the properties of the resulting image. If the object is placed beyond the focal length of a converging lens, a real, inverted image is formed. However, if the object is within the focal length, the lens will form a virtual, upright image.

The concept of object distance plays a role in everyday applications, such as setting up a projector for clear image display or adjusting the focus on a camera. For any lens system, if we know two of the three variables (object distance, image distance, and focal length), we can easily calculate the third.
Image Distance
The image distance, represented as \(d_i\), is the distance from the converging lens to the image that forms as a result of the refraction of light through the lens. The sign of the image distance also informs us of the nature of the image: positive for real images and negative for virtual images. Calculating \(d_i\) is integral in designing optical systems, as this determines where an image is formed and how it can be captured or viewed.

In scenarios where the object distance is known, along with the lens's focal length, the image distance can be calculated using the lens formula, allowing for the prediction of where an image will form in relation to the lens. This prediction is critical in applications such as the focusing mechanism of cameras or the design of eyeglasses, where the correct image placement is paramount for functionality.
Focal Length
The focal length, denoted by \(f\), is one of the fundamental properties of a lens, representing the distance between the lens and its focal point. It essentially measures how strongly the lens converges or diverges light. For a converging lens, the focal length is positive, indicating that the convergent point is real and on the opposite side of the lens.

Understanding focal length is key to mastering concepts in optics, as it affects the magnification, image size, and clarity in optical instruments. The shorter the focal length, the more powerful the lens is at bending light, resulting in greater magnification. Conversely, a longer focal length implies less curvature and weaker converging power.

In the lens formula, the focal length is factored into determining both the object distance and image distance. It is a fixed property for a given lens and is essential for tasks like calculating the necessary lens power in corrective eyewear or setting up optical systems in photography and astronomy.

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Most popular questions from this chapter

You are given a thin converging lens with a focal length of \(10 \mathrm{~cm}\). a) An object is placed at \(6 \mathrm{~cm}\) from the lens. Draw a ray diagram for the situation. b) Calculate the image distance. Does the calculated distance agree with the distance given by the ray diagram? Is the image real or virtual? c) The object is moved out to \(8 \mathrm{~cm}\). Draw a ray diagram for the situation. d) Calculate the image distance. Does the calculated distance agree with the distance given by the ray diagram? Is the image real or virtual. Is it larger or smaller than the image from part (a)?

A beam of unpolarized light is passed through two polarizers. If the polarization axis of the second polarizer is crossed at an angle of \(90^{\circ}\) with respect to the axis of the first polarizer, then the intensity of light seen by someone located to the right of the second polarizer is (A) the intensity of the original light. (B) one-half the intensity of the original light. (C) one-quarter the intensity of the original light. (D) one-eighth the intensity of the original light. (E) zero.

A beam of parallel light rays is incident on the same screen in the previous problem. If you want to produce a diffraction pattern on the distant wall, you should place in the clamps which of the following optical devices? (A) A prism (B) A narrow slit in a piece of cardboard (C) A rectangular piece of glass (D) A converging lens (E) A diverging lens

A beam of unpolarized light is passed through two polarizers. If the polarization axis of the second polarizer is at an angle of \(90^{\circ}\) with respect to the axis of the first polarizer, then the intensity of light seen by someone located between them is (A) the intensity of the original light. (B) one-half the intensity of the original light. (C) one-quarter the intensity of the original light. (D) one-eighth the intensity of the original light. (E) zero.

A beam of parallel light rays is incident on an opaque screen that has a large opening in it. The opening is fitted with clamps designed to hold optical devices. If you want to produce a tiny spot of light on a distant wall, you should place in the clamps which of the following devices? (A) A prism (B) A narrow slit in a piece of cardboard (C) A rectangular piece of glass (D) A converging lens (E) A diverging lens

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