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The distance between an object and its upright image is \(d\). If the magnification is \(M,\) what is the focal length of the lens that is being used to form the image?

Short Answer

Expert verified
The focal length of the lens is \(-\(\frac{1}{M + 1}\)\)

Step by step solution

01

Express Image and Object Distance in Terms of Given Distance

Since the distance \(d\) between the object and image is given as \(|v - u|\), if the image is located to the right of the object, then \(v = d + u\)
02

Substitute Image Distance into Magnification Formula

Next, Substituting for \(v\) from step 1 into the magnification formula \(-\(\frac{v}{u}\)\), it transforms to \(M = -\(\frac{d + u}{u}\) = -1 - \(\frac{d}{u}\)\)
03

Express Object Distance in Terms of Magnification

Rearranging the formula from Step 2 to solve for \(u\), we have \(u = -\(\frac{d}{M+1}\)\)
04

Substitute into Mirror Lens Formula and Solve for Focal Length

Substituting for \(v\) and \(u\) into the mirror lens formula gives \(\frac{1}{f} = \(\frac{1}{d + u}\) + \(\frac{1}{u}\) = \(\frac{1}{d - \(\frac{d}{M+1}\)}\) + \(-\(\frac{M+1}{d}\)\), which simplifies to \(\frac{1}{f} = -M - 1\)\. Therefore, the focal length \(f = -\(\frac{1}{M + 1}\)\)

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Key Concepts

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

Magnification
Magnification in optics refers to how much larger or smaller an image appears compared to the object itself. It is a ratio that describes the change in size. To understand magnification, consider that it is defined as the ratio of the image distance (\(v\)) to the object distance (\(u\)), given by the formula:
  • \( M = -\frac{v}{u} \)
Here, the negative sign indicates that the image formed by the lens is inverted compared to the object. When a magnification (\( M \)) is mentioned, it often tells you how much the image size varies with respect to the object. For instance, if the magnification is -2, the image is twice as large as the original object and inverted. Understanding magnification helps us predict the scale of images and is crucial when designing optical systems that rely on lenses, such as microscopes and telescopes.
Lens Formula
The lens formula is a critical equation in optics that relates the focal length of a lens to the object distance and the image distance. The formula is given as:
  • \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \)
This fundamental equation allows one to calculate any one of the three parameters if the other two are known. Here, \(f\) is the focal length, \(v\) is the image distance, and \(u\) is the object distance. The formula originates from the geometric principles of how light bends when it passes through the lens. By manipulating this formula, it becomes possible to predict where an image will form and how lenses can be designed for various applications, from eyeglasses to cameras.
The lens formula is particularly useful when working with convex or concave lenses, enabling a precise setup for clear, focused images.
Object Distance
Object distance in the context of optics refers to how far the object is from the lens. It is denoted by \(u\) in equations. The object distance greatly influences where and how the image will form when passing through a lens. In practice, object distance helps us determine placement in relation to optical components for achieving desired image characteristics.
Moreover, through equations like the lens formula and magnification formula, object distance is crucial in calculating image properties and focal length. As seen in solving the lens equation, modifying object distance can manipulate the magnification and affect the convergence or divergence of light creating the image. In simple terms, knowing the object distance lets you tailor the optical arrangement, crucial for tasks ranging from photography to experimental physics setups.

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Most popular questions from this chapter

A cataract-impaired lens in an eye may be surgically removed and replaced by a manufactured lens. The focal length required for the new lens is determined by the lens-to-retina distance, which is measured by a sonar-like device, and by the requirement that the implant provide for correct distant vision. (a) Assuming the distance from lens to retina is \(22.4 \mathrm{mm},\) calculate the power of the implanted lens in adopters. (b) Because no accommodation occurs and the implant allows for correct distant vision, a corrective lens for close work or reading must be used. Assume a reading distance of \(33.0 \mathrm{cm}\) and calculate the power of the lens in the reading glasses.

A flint glass plate \((n=1.66)\) rests on the bottom of an aquarium tank. The plate is \(8.00 \mathrm{cm}\) thick (vertical dimension) and is covered with a layer of water \((n=1.33)\) \(12.0 \mathrm{cm}\) deep. Calculate the apparent thickness of the plate as viewed from straight above the water.

Suppose an object has thickness \(d p\) so that it extends from object distance \(p\) to \(p+d p .\) Prove that the thickness \(d q\) of its image is given by \(\left(-q^{2} / p^{2}\right) d p,\) so that the longitudinal \(M\) is the lateral magnification \(d q / d p=-M^{2},\) where magnification.

An object located \(32.0 \mathrm{cm}\) in front of a lens forms an image on a screen \(8.00 \mathrm{cm}\) behind the lens. (a) Find the focal length of the lens. (b) Determine the magnification. (c) Is the lens converging or diverging?

In a church choir loft, two parallel walls are \(5.30 \mathrm{m}\) apart. The singers stand against the north wall. The organist faces the south wall, sitting \(0.800 \mathrm{m}\) away from it. To enable her to see the choir, a flat mirror \(0.600 \mathrm{m}\) wide is mounted on the south wall, straight in front of her. What width of the north wall can she see? Suggestion: Draw a topview diagram to justify your answer.

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