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A lens forms a real image that is \(214 \mathrm{~cm}\) away from the object and \(1 \frac{2}{3}\) times its height. What kind of lens is this, and what is its focal length?

Short Answer

Expert verified
The lens is a converging lens and its focal length is 214 cm.

Step by step solution

01

Identifying the Lens

The lens forms a real image and bigger than the object, which indicates it is a converging or a positive lens.
02

Find Object Distance (u)

The image is \(1 \frac{2}{3}\) times the height of the object. This is the magnification (m = -v/u). In this case, m is \(1 \frac{2}{3}\) or 5/3 . Also from the problem, the image distance (v) is -214 cm because it is real and on the opposite side of the lens. Therefore, the object distance is u = -v/m = -{-214 cm}/{5/3} = -128.4 cm. Note that u is negative because it is on the opposite side of the image based on the sign convention.
03

Find Focal Length (f)

Now we can use the lens formula to find the focal length of the lens: 1/f = 1/v - 1/u, substituting v = -214 cm and u = -128.4 cm: 1/f = 1/(-214) - 1/(-128.4), which simplifies to: 1/f = -0.00467cm^{-1}, so the focal length f = -214 cm.

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Key Concepts

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

Real Image
A real image is created when light rays actually converge at a point after passing through a lens. Unlike a virtual image, a real image can be captured on a screen because the light physically reaches that point. You find real images commonly in projectors and cameras.
A key distinction of real images is that they are always formed on the opposite side of the lens from the object. This occurs when the object is placed outside the focal point of a converging lens. Real images are typically inverted compared to the object.
In the context of this exercise, the lens forms a real image 214 cm away from the object, indicating a genuine convergence of light rays.
Converging Lens
Converging lenses, also known as convex lenses, have a shape that bulges outward. These lenses focus incoming parallel light rays to a point known as the focal point. This property makes them indispensable in applications like magnifying glasses and eyeglasses for farsightedness.
  • Converging lenses can form both real and virtual images based on the object's position relative to the focal point.
  • When an object is beyond the focal length, they produce real and inverted images.
  • If the object is placed within the focal length, a virtual and magnified image appears.
In our exercise, the lens is identified as a converging lens as it forms a real image and magnifies the object.
Magnification
Magnification is the factor by which an image's size changes relative to the object. It is determined by the formula:\[m = \frac{-v}{u}\]where \(m\) is the magnification, \(v\) is the image distance, and \(u\) is the object distance.
A magnification value greater than one means the image is larger than the object, while a magnification less than one indicates a smaller image. Negative magnification denotes that the image is inverted compared to the object.
For this scenario, the image is \(1\frac{2}{3}\) times larger than the object, indicating a magnification of \(\frac{5}{3}\). This positive value means the image is bigger and is located on the opposite side of the lens from the object. As calculated in the solution, this aligns with the behavior of real images produced by converging lenses.
Focal Length
The focal length of a lens is the distance from the lens to its focal point. It determines how strongly the lens converges or diverges light. The formula to find the focal length using object and image distances is given by the lens equation:\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]where \(f\) is the focal length, \(v\) is the image distance, and \(u\) is the object distance.
For converging lenses, the focal length is positive, indicating that they focus light. In our exercise, applying the lens formula with \(v = -214\) cm and \(u = -128.4\) cm allows us to solve for \(f\). This computation yields a focal length demonstrating the lens's capability to focus light into a real image.

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Most popular questions from this chapter

A transparent rod \(30.0 \mathrm{~cm}\) long is cut flat at one end and rounded to a hemispherical surface of radius \(10.0 \mathrm{~cm}\) at the other end. A small object is embedded within the rod along its axis and halfway between its ends, \(15.0 \mathrm{~cm}\) from the flat end and \(15.0 \mathrm{~cm}\) from the vertex of the curved end. When the rod is viewed from its flat end, the apparent depth of the object is \(8.20 \mathrm{~cm}\) from the flat end. What is its apparent depth when the rod is viewed from its curved end?

An object with height \(4.00 \mathrm{~mm}\) is placed \(28.0 \mathrm{~cm}\) to the left of a converging lens that has focal length \(8.40 \mathrm{~cm} .\) A second lens is placed \(8.00 \mathrm{~cm}\) to the right of the converging lens. (a) What is the focal length of the second lens if the final image is inverted relative to the \(4.00-\mathrm{mm}\) -tall object and has height \(5.60 \mathrm{~mm} ?\) (b) What is the distance between the original object and the final image?

A person is lying on a diving board \(3.00 \mathrm{~m}\) above the surface of the water in a swimming pool. She looks at a penny that is on the bottom of the pool directly below her. To her, the penny appears to be a distance of \(7.00 \mathrm{~m}\) from her. What is the depth of the water at this point?

The radii of curvature of the surfaces of a thin converging meniscus lens are \(R_{1}=+12.0 \mathrm{~cm}\) and \(R_{2}=+28.0 \mathrm{~cm} .\) The index of refraction is \(1.60 .\) (a) Compute the position and size of the image of an object in the form of an arrow \(5.00 \mathrm{~mm}\) tall, perpendicular to the lens axis, \(45.0 \mathrm{~cm}\) to the left of the lens. (b) A second converging lens with the same focal length is placed \(3.15 \mathrm{~m}\) to the right of the first. Find the position and size of the final image. Is the final image erect or inverted with respect to the original object? (c) Repeat part (b) except with the second lens \(45.0 \mathrm{~cm}\) to the right of the first.

A frog can see an insect clearly at a distance of \(10 \mathrm{~cm}\). At that point the effective distance from the lens to the retina is \(8 \mathrm{~mm} .\) If the insect moves \(5 \mathrm{~cm}\) farther from the frog, by how much and in which direction does the lens of the frog's eye have to move to keep the insect in focus? (a) \(0.02 \mathrm{~cm}\), toward the retina; (b) \(0.02 \mathrm{~cm}\), away from the retina; (c) \(0.06 \mathrm{~cm}\), toward the retina; (d) \(0.06 \mathrm{~cm}\), away from the retina.

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