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A projection lens is employed to produce \(2.4 \mathrm{~m} \times 3.2 \mathrm{~m}\) pictures from \(3.0 \mathrm{~cm} \times 4.0 \mathrm{~cm}\) slides on a screen that is \(25 \mathrm{~cm}\) from the lens. Compute its focal length.

Short Answer

Expert verified
The focal length of the lens is approximately 0.31 cm.

Step by step solution

01

Understand the Problem

We have a lens projecting images from slides onto a screen. The dimensions of the slide are 3.0 cm by 4.0 cm, and they are projected to a 2.4 m by 3.2 m image on a screen 25 cm away from the lens. We need to find the focal length of the lens.
02

Calculate the Magnification

The magnification \(M\) of the lens can be determined by comparing the dimensions of the image to the slide image. Given the image dimensions 2.4 m \(= 240 \, \text{cm}\) and 3.2 m \(= 320 \, \text{cm}\) and the slide dimensions 3.0 cm and 4.0 cm, the magnifications are \(M_{width} = \frac{240}{3} = 80\) and \(M_{height} = \frac{320}{4} = 80\). Both dimensions have the same magnification of 80.
03

Relate Magnification to Distances

The magnification \(M\) is given by \(M = \frac{v}{u}\), where \(v\) is the image distance (25 cm) and \(u\) is the object distance. Since \(M = 80\), rearrange the formula to find \(u\): \(80 = \frac{25}{u}\). Solving for \(u\), we get \(u = \frac{25}{80} = 0.3125 \, \text{cm}\).
04

Use the Lens Formula

The lens formula is given by \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \). Substitute \(v = 25 \, \text{cm}\) and \(u = 0.3125 \, \text{cm}\) into the equation: \( \frac{1}{f} = \frac{1}{25} + \frac{1}{0.3125} \).
05

Calculate the Focal Length

Calculate \( \frac{1}{u} = \frac{1}{0.3125} = 3.2 \) and \( \frac{1}{v} = \frac{1}{25} = 0.04 \), which gives \( \frac{1}{f} = 0.04 + 3.2 \). Thus, \( \frac{1}{f} = 3.24 \), and the focal length is \( f = \frac{1}{3.24} \approx 0.3099 \, \text{cm} \).
06

Review the Solution

Verify the calculation steps and units. All values are correctly used, confirming a focal length of approximately 0.31 cm.

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Key Concepts

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

Lens Formula
The lens formula is a fundamental equation in optics that helps us relate the focal length of a lens to the distances of the object and the image from the lens. Mathematically, it is expressed as:\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]where:- \( f \) is the focal length of the lens,- \( v \) is the image distance (the distance from the lens to the image),- \( u \) is the object distance (the distance from the lens to the object).
This formula is quite versatile and is used for both converging and diverging lenses. In our exercise, we needed to find the focal length \( f \) of a projection lens. By applying the given values for \( v \) and \( u \), which were calculated using the magnification, we could solve for \( f \). It’s important to ensure that all distances are in the same units before plugging them into the formula. This will help avoid calculation errors. Always verify calculations to ensure accuracy in optics.
Magnification in Optics
Magnification is a key concept in optics that describes how much larger or smaller the image is compared to the object. The magnification \( M \) is defined by the ratio of the image size to the object size, which can also be expressed in terms of distances as:\[ M = \frac{v}{u} \]This relationship tells us that the magnification of a lens depends on the ratio of the image distance \( v \) to the object distance \( u \).
In the exercise, we calculated the magnification by comparing the dimensions of the image on a screen to the dimensions of the slide. Both dimensions provided a magnification value of 80. This consistent magnification factor can be used in applications where uniform scaling is necessary, like in projection systems.
Knowing the magnification helped us find the object distance \( u \) from the lens, which is crucial for further calculations with the lens formula. Understanding how to compute and apply magnification ensures that images are correctly sized in various optical devices.
Projection Lens
A projection lens is a specific type of lens used to project an enlarged or diminished image from one surface to another. Often used in projectors, this lens takes a small image and casts a much larger one onto a screen.
Key characteristics of a projection lens include the focal length, which determines how far the lens can project an image before it becomes blurry, and the magnification, which indicates how large the projected image will be compared to the original.
In this exercise, the goal was to compute the focal length of such a lens based on given dimensions of the slide and the projected image. The logic behind using magnification and lens formula supports the design and application of projection lenses in real-world scenarios, such as in cinema projectors or overhead projective lenses. These lenses are crucial for systems where precise image projection is needed over varying distances.

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Most popular questions from this chapter

Reflecting telescopes make use of a concave mirror, in place of the objective lens, to bring the distant object into focus. What is the magnifying power of a telescope that has a mirror with \(250 \mathrm{~cm}\) radius and an eyepiece whose focal length is \(5.0 \mathrm{~cm}\) ? As it is for a refracting telescope (i.e., one with two lenses), \(M_{A}=\) \(-f_{O} / f_{E}\) again applies where, in this case, \(f_{O}=-R / 2=125 \mathrm{~cm}\) and \(f_{E}\) \(=5.0 \mathrm{~cm} .\) Thus, \(M_{A}=-25\).

A lens combination consists of two lenses with focal lengths of \(+4.0 \mathrm{~cm}\) and \(+8.0 \mathrm{~cm}\), which are spaced \(16 \mathrm{~cm}\) apart. Locate and describe the image of an object placed \(12 \mathrm{~cm}\) in front of the \(+4.0\) cm lens.

An astronomical telescope with an objective lens of focal length \(+80 \mathrm{~cm}\) is focused on the moon. By how much must the eyepiece be moved to focus the telescope on an object 40 meters distant?

The large telescope at Mt. Palomar has a concave objective mirror diameter of \(5.0 \mathrm{~m}\) and radius of curvature \(46 \mathrm{~m}\). What is the magnifying power of the instrument when it is used with an eyepiece of focal length \(1.25 \mathrm{~cm}\) ?

In a compound microscope, the focal lengths of the objective and eyepiece are \(+0.50 \mathrm{~cm}\) and \(+2.0 \mathrm{~cm}\), respectively. The instrument is focused on an object \(0.52 \mathrm{~cm}\) from the objective lens. Compute the magnifying power of the microscope if the virtual image is viewed by the eye at a distance of \(25 \mathrm{~cm}\).

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