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(a) You want to use a lens with a focal length of \(35.0 \mathrm{~cm}\) to produce a real image of an object, with the image twice as long as the object itself. What kind of lens do you need, and where should the object be placed? (b) Suppose you want a virtual image of the same object, with the same magnification-what kind of lens do you need, and where should the object be placed?

Short Answer

Expert verified
(a) Use a convex lens with the object 70 cm from the lens. (b) Use a convex lens with the object 17.5 cm from the lens.

Step by step solution

01

Identifying the Lens Type for a Real Image

A real image is formed when light actually converges and passes through the image point. This can only happen with a **converging lens**, also known as a convex lens. Thus, we need a convex lens for part (a).
02

Using the Lens Formula for a Real Image

The lens formula is given by: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance. Substituting \( f = 35.0 \text{ cm} \), we need to determine \( u \) and \( v \).
03

Using Magnification Formula for Real Image

The magnification \( m \) is defined as \( m = \frac{h_i}{h_o} = \frac{v}{u} = 2 \), since the image is twice the length of the object. Thus, \( v = 2u \).
04

Solving for Object Distance (Real Image)

Substitute \( v = 2u \) into the lens formula:\[ \frac{1}{35} = \frac{1}{2u} - \frac{1}{u} \]This simplifies to:\[ \frac{1}{35} = -\frac{1}{2u} \]Solving gives \( u = 70 \text{ cm} \). Therefore, the object should be placed 70 cm from the lens.
05

Identifying the Lens Type for a Virtual Image

A virtual image is formed when light does not actually converge, but appears to diverge from a point. This condition can occur with a **diverging lens** (concave) or a **converging lens** if the object is inside the focal point. For a virtual image with a positive magnification, we use a **converging (convex) lens** with the object inside the focal length.
06

Using the Lens Formula for a Virtual Image

For the same magnification of \( 2 \), the magnification formula is similarly \( m = -\frac{v}{u} = 2 \). Since the image is virtual, \( v < 0 \) (since it diverges from a point). We now use the formula:\[ -v = 2u \]
07

Solving for Object Distance (Virtual Image)

Substituting back into the lens formula:\[ \frac{1}{35} = \frac{-1}{2u} - \frac{1}{u} \]This simplifies to:\[ \frac{1}{35} = -\frac{3}{2u} \]Solving gives \( u = \frac{105}{2} \text{ cm} \approx 17.5 \text{ cm} \). Therefore, the object should be placed approximately 17.5 cm from the lens.

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Key Concepts

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

converging lens
A converging lens, also known as a convex lens, is an optical lens that brings parallel rays of light to a focus. This means that all the light entering through the lens gets bent towards a central point, known as the focal point. Converging lenses are thicker in the middle than at the edges. They are commonly used in various applications because they can form real images by actually focusing light. For instance, they are used in cameras, eyeglasses for farsightedness, and of course, in experiments and studies involving the principles of optics.
When using a converging lens, it's crucial to understand the focal length, as it determines how strongly the lens converges or focuses light. If the focal length is long, the lens is weaker and bends light less sharply. Conversely, a shorter focal length indicates a stronger lens that focuses light more sharply. In practical applications, the focal length guides the positioning of the object and the expectation of the image distance.
real image
A real image occurs when light rays actually converge at a point after passing through a lens. Because they truly come together, real images can be projected onto a screen. This is one of the hallmark features that differentiate real images from virtual ones.
Real images created by converging lenses are always inverted, meaning they appear upside down compared to the object. They can vary in size depending on the lens configuration. Since light converges, the focal point is reached on the opposite side of the lens from the object. The position of a real image is crucially dependent on the distance between the object and the lens. The formula that relates to this positioning is called the lens formula, and it's an essential tool for calculating configurations:
  • Real images have positive image distances in lens calculations.
  • The image distance (\( v \)) becomes more accessible with clear understanding and calculation via lens formula.
Identifying and projecting a real image requires carefully placing the object beyond the focal length of a converging lens.
virtual image
Unlike real images, virtual images are formed when light rays appear to originate from a point. However, they don't actually come together. Instead, virtual images exist only as the visual perception of an observer looking through a lens. That's why they cannot be projected on a screen.
Virtual images are formed when the distance of the object is shorter than the focal length, primarily in converging lenses when the object resides inside the focal point. To distinguish a virtual image:
  • Virtual images always appear upright in comparison to the original object.
  • They have negative image distances, a characteristic that comes into play in the lens formula.
  • The magnification formula shows positive values, indicating the image is upright compared to the object.
Most commonly used in devices like magnifying glasses, virtual images play a vital role where an enlarged and upright perception is necessary.
magnification formula
The magnification formula is a mathematical expression used to determine how much larger or smaller an image is compared to the object itself. In optics, the formula is defined as:\[ m = \frac{h_i}{h_o} = \frac{v}{u} \]where \( m \) is magnification, \( h_i \) is the height of the image, \( h_o \) is the height of the object, \( v \) is the image distance, and \( u \) is the object distance.
This formula is pivotal for calculations involving both real and virtual images:
  • A positive magnification signifies an upright image. This is typically a feature of virtual images.
  • A negative magnification signifies an inverted image, which is common for real images.
  • The absolute value of the magnification tells you how many times larger or smaller the image is relative to the object.
Understanding and using the magnification formula allows us to achieve desired image sizes and orientations by expertly positioning the object and choosing appropriate lenses.

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Most popular questions from this chapter

A spherical, concave shaving mirror has a radius of curvature of \(32.0 \mathrm{~cm} .\) (a) What is the magnification of a person's face when it is \(12.0 \mathrm{~cm}\) to the left of the vertex of the mirror? (b) Where is the image? Is the image real or virtual? (c) Draw a principal-ray diagram showing the formation of the image.

The focal length of a mirror can be determined entirely from the shape of the mirror. In contrast, to determine the focal length of a lens, we must know both the shape of the lens and its index of refraction - as well as the index of refraction of the surrounding medium. For instance, when a thin lens is immersed in a liquid, we must modify the thin-lens equation to take into account the refractive properties of the surrounding liquid: $$\frac{1}{f}=\left(\frac{n}{n_{\mathrm{liq}}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right),$$ where \(n_{\text {liq }}\) is the index of refraction of the liquid and \(n\) is the index of refraction of the glass. In air, a particular person sees clearly when wearing eyeglasses that have glass lenses with a specific radius of curvature. To see clearly when scuba diving, this person wants to have her optometrist make new lenses that can be mounted in front of her scuba mask so that the lenses are surrounded by water. Which of the following is true if the new lenses are to have the same focal length in water as the original lenses do in air? The new lenses need to A. have a larger radius of curvature compared to the original lenses. B. have a smaller radius of curvature compared to the original lenses. C. have the same radius of curvature as the original lenses. D. be flat pieces of glass.

Combination of lenses, I. When two lenses are used in combination, the first one forms an image that then serves as the object for the second lens. The magnification of the combination is the ratio of the height of the final image to the height of the object. A \(1.20-\mathrm{cm}-\) tall object is \(50.0 \mathrm{~cm}\) to the left of a converging lens of focal length \(40.0 \mathrm{~cm} .\) A second converging lens, this one having a focal length of \(60.0 \mathrm{~cm},\) is located \(300.0 \mathrm{~cm}\) to the right of the first lens along the same optic axis. (a) Find the location and height of the image (call it \(I_{1}\) ) formed by the lens with a focal length of \(40.0 \mathrm{~cm}\). (b) \(I_{1}\). is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses.

The left end of a long glass rod \(6.00 \mathrm{~cm}\) in diameter has a convex hemispherical surface \(3.00 \mathrm{~cm}\) in radius. The refractive index of the glass is \(1.60 .\) Determine the position of the image if an object is placed in air on the axis of the rod at the following distances to the left of the vertex of the curved end: (a) infinitely far, (b) \(12.0 \mathrm{~cm},\) and (c) \(2.00 \mathrm{~cm}\)

The two surfaces of a plastic converging lens have equal radii of curvature of \(22.0 \mathrm{~cm},\) and the lens has a focal length of \(20.0 \mathrm{~cm} .\) Calculate the index of refraction of the plastic.

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