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Chapter 18: Problem 46

Solve the given applied problems involving variation. The \(f\) -number lens setting of a camera varies directly as the square root of the time \(t\) that the film is exposed. If the \(f\) -number is 8 (written as \(f / 8\) ) for \(t=0.0200\) s, find the \(f\) -number for \(t=0.0098 \mathrm{s}\)

Short Answer

Expert verified
The \( f \)-number for \( t = 0.0098 \, ext{s} \) is approximately 5.60.

Step by step solution

01

Understanding Direct Variation

First, we need to set up a direct variation equation to solve the problem. The relationship is given by the equation: \[ f = k \sqrt{t} \]where \( f \) is the \( f \)-number, \( k \) is the constant of variation, and \( t \) is the time.
02

Find the Constant of Variation

We know \( f = 8 \) when \( t = 0.0200 \, ext{s} \). Plugging these into the equation, we get:\[ 8 = k \sqrt{0.0200} \]Solving for \( k \), we have:\[ k = \frac{8}{\sqrt{0.0200}} \]
03

Calculate the Constant

Calculate the square root of 0.0200 and then calculate \( k \).\[ \sqrt{0.0200} = 0.1414 \]\[ k = \frac{8}{0.1414} \approx 56.57 \]So the constant \( k \) is approximately 56.57.
04

Find the New \( f \)-Number

Now, use the constant \( k \) to find the \( f \)-number for \( t = 0.0098 \, ext{s} \):\[ f = 56.57 \times \sqrt{0.0098} \]
05

Calculate the New \( f \)-Number

Calculate the square root of 0.0098 and then find \( f \).\[ \sqrt{0.0098} = 0.0989 \]\[ f = 56.57 \times 0.0989 \approx 5.60 \]Thus, the \( f \)-number for \( t = 0.0098 \, ext{s} \) is approximately 5.60.

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Key Concepts

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

Understanding the f-number
The f-number, also known as the f-stop, is an important setting in photography that controls the aperture of a camera's lens. The aperture refers to the opening in the lens through which light travels. The f-number determines how wide this opening is.
  • A lower f-number means a wider opening, allowing more light to enter, which is useful in low-light conditions.
  • A higher f-number results in a narrower opening, reducing the amount of light but increasing the depth of field, making more of the scene sharp and in focus.
The f-number is represented as something like f/2.8 or f/8, where the number indicates the ratio of the lens's focal length to the diameter of the aperture. Therefore, the f-number not only affects the exposure but also the overall composition and clarity of the photograph.
What is Exposure Time?
Exposure time, or shutter speed, is the duration for which the camera's shutter remains open to allow light to hit the sensor or film. This setting significantly affects how motion is represented in photographs.
  • Short exposure times (e.g., 1/1000 of a second) freeze fast-moving subjects, capturing them in detail.
  • Long exposure times (e.g., several seconds) can blur moving objects, creating effects like soft, flowing waterfalls or streaky city lights at night.
Exposure time works in conjunction with the f-number to control the light exposure on the image sensor. By adjusting these two settings, photographers can achieve different artistic effects and also ensure that photos are not too bright (overexposed) or too dark (underexposed). Together, they form part of the essential triangle in photography composition known as the exposure triangle, which also includes ISO settings.
Exploring the Constant of Variation
In context of this problem, the constant of variation, represented by \( k \), plays a crucial role in the relationship between the f-number and exposure time. Direct variation describes how one quantity directly impacts another, and in this problem, it means that the f-number changes with the square root of the exposure time. When the problem states "varies directly as the square root," it indicates a mathematical relationship that can be expressed with the equation \( f = k \sqrt{t} \). Here, \( f \) is the f-number, \( t \) is the exposure time, and \( k \) is a constant.
  • The constant \( k \) is determined once specific values of \( f \) and \( t \) are known. In this example, \( k \approx 56.57 \) was derived when \( f = 8 \) and \( t = 0.0200 \) seconds.
  • To find \( f \) at a new exposure time, this constant remains unchanged, highlighting its role as a proportionality factor.
Understanding this constant helps in adapting camera settings under different lighting conditions, ensuring high-quality photography with correct exposures.

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Most popular questions from this chapter

Solve the given applied problems involving variation. The electric resistance \(R\) of a wire varies directly as its length \(l\) and inversely as its cross-sectional area \(A\). Find the relation between resistance, length, and area for a wire that has a resistance of \(0.200 \Omega\) for a length of \(225 \mathrm{ft}\) and crosssectional area of 0.0500 in. \(^{2}\)The general gas law states that the pressure \(P\) of an ideal gas varies directly as the thermodynamic temperature \(T\) and inversely as the volume \(V\). If \(P=610 \mathrm{kPa}\) for \(V=10.0 \mathrm{cm}^{3}\) and \(T=290 \mathrm{K},\) find \(V\) for \(P=400 \mathrm{kPa}\) and \(T=400 \mathrm{K}\)

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Solve the given applied problems involving variation. The escape velocity \(v\) a spacecraft needs to leave the gravitational field of a planet varies directly as the square root of the product of the planet's radius \(R\) and its acceleration due to gravity \(g .\) For Mars and earth, \(R_{M}=0.533 R_{e}\) and \(g_{M}=0.400 g_{e^{*}}\) Find \(v_{M}\) for Mars if \(v_{e}=11.2 \mathrm{km} / \mathrm{s}\)

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Find the required quantities from the given proportions. For two connected gears, the relation \(\frac{d_{1}}{d_{2}}=\frac{N_{1}}{N_{2}}\) holds, where \(d\) is the diameter of the gear and \(N\) is the number of teeth. Find \(N_{1}\) if \(d_{1}=2.60\) in. \(d_{2}=11.7\) in. and \(N_{2}=45 .\) The ratio \(N_{2} / \overline{N_{1}}\) is called the gear ratio. See Fig. \(18.3 .\)
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