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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

When an object is placed at the proper distance to the left of a converging lens, the image is focused on a screen \(30.0 \mathrm{~cm}\) to the right of the lens. A diverging lens is now placed \(15.0 \mathrm{~cm}\) to the right of the converging lens, and it is found that the screen must be moved \(19.2 \mathrm{~cm}\) farther to the right to obtain a sharp image. What is the focal length of the diverging lens?

The image of a tree just covers the length of a plane mirror \(4.00 \mathrm{~cm}\) tall when the mirror is held \(35.0 \mathrm{~cm}\) from the eye. The tree is \(28.0 \mathrm{~m}\) from the mirror. What is its height?

The overall angular magnification of a microscope is \(M=-178 .\) The eyepiece has focal length \(15.0 \mathrm{~mm}\) and the final image is at infinity. The separation between the two lenses is \(202 \mathrm{~mm}\). What is the focal length of the objective? Do not use the approximation \(s_{1} \approx f_{1}\) in the expression for \(M\).

A compound microscope has an objective lens with focal length \(14.0 \mathrm{~mm}\) and an eyepiece with focal length \(20.0 \mathrm{~mm}\). The final image is at infinity. The object to be viewed is placed \(2.0 \mathrm{~mm}\) beyond the focal point of the objective lens. (a) What is the distance between the two lenses? (b) Without making the approximation \(s_{1} \approx f_{1},\) use \(M=m_{1} M_{2}\) with \(m_{1}=-s_{1}^{\prime} / s_{1}\) to find the overall angular magnification of the microscope. (c) What is the percentage difference between your result and the result obtained if the approximation \(s_{1} \approx f_{1}\) is used to find \(M ?\)

(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 height of the image twice the height of the object. 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?

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