Problem 100
Question
The equation \(\frac{1}{p}+\frac{1}{q}=\frac{1}{f}\) relates an objects distance \(p\) from a lens and the image distance \(q\) from the lens to the focal length \(f\) of the lens. (a) Determine the focal length of a lens in which an object 8 feet away produces an image 6 feet away. (b) Determine how far an object is from a lens if the focal length of the lens is 10 inches and the image distance is 10 inches. (c) Determine how far an object will be from a lens that has a focal length of \(1 \frac{7}{8} \mathrm{~cm}\) and the object distance is \(3 \mathrm{~cm}\) away from the lens.
Step-by-Step Solution
Verified Answer
Question: Based on the given equation \(\frac{1}{p}+\frac{1}{q}=\frac{1}{f}\), determine the image distance (q) if the focal length (f) is \(\frac{15}{8}\mathrm{cm}\) and the object distance (p) is 3cm.
Answer: The image distance (q) is 5 cm.
1Step 1: (a) Determine the focal length of the lens.
First, we can find the focal length of the lens using the given equation:
\(\frac{1}{p}+\frac{1}{q}=\frac{1}{f}\). Plug in the given object distance \(p=8\mathrm{ft}\) and image distance \(q=6\mathrm{ft}\):
\(\frac{1}{8}+\frac{1}{6}=\frac{1}{f}\). To solve for the focal length f, first find a common denominator for the fractions on the left side:
\(\frac{3}{24}+\frac{4}{24}=\frac{1}{f}\). Adding the fractions:
\(\frac{7}{24}=\frac{1}{f}\). Finally, take the reciprocal of both sides to find f:
\(f=\frac{24}{7} \approx 3.43\mathrm{ft}\).
2Step 2: (b) Determine the object distance (p).
Use the given equation \(\frac{1}{p}+\frac{1}{q}=\frac{1}{f}\) with the given focal length \(f=10\mathrm{in}\) and image distance \(q=10\mathrm{in}\):
\(\frac{1}{p}+\frac{1}{10}=\frac{1}{10}\). Subtract \(\frac{1}{10}\) from both sides to isolate \(\frac{1}{p}\):
\(\frac{1}{p}=0\). Since it is not possible for \(\frac{1}{p}\) to be zero, this situation is not physically possible—a lens with a focal length of 10 inches cannot produce an image distance of 10 inches.
3Step 3: (c) Determine the image distance (q).
Use the provided equation \(\frac{1}{p}+\frac{1}{q}=\frac{1}{f}\) with the given focal length \(f=\frac{15}{8}\mathrm{cm}\) and object distance \(p=3\mathrm{cm}\):
\(\frac{1}{3}+\frac{1}{q}=\frac{8}{15}\). To solve for q, first subtract \(\frac{1}{3}\) from both sides:
\(\frac{1}{q}=\frac{8}{15}-\frac{1}{3}\). Find a common denominator and subtract the fractions:
\(\frac{1}{q}=\frac{8}{15}-\frac{5}{15}\). Simplify the right side:
\(\frac{1}{q}=\frac{3}{15}\). Take the reciprocal of both sides to find q:
\(q=\frac{15}{3}=5\mathrm{cm}\). The object will be 5 cm away from the lens.
Key Concepts
Algebraic Fraction OperationsReciprocal of FractionsLens Formula Application
Algebraic Fraction Operations
Understanding algebraic fraction operations is crucial when working with equations involving fractions, such as the lens formula. Algebraic fractions are expressions that contain fractions with variables, like the ones seen in the equation \( \frac{1}{p} + \frac{1}{q} = \frac{1}{f} \). When working with algebraic fractions, the first goal is to combine them by finding a common denominator, which allows for the addition or subtraction of the fractions. For instance, in the calculation of the focal length \(f\) when \(p = 8\) feet and \(q = 6\) feet, we converted \( \frac{1}{8} + \frac{1}{6}\) into \( \frac{3}{24} + \frac{4}{24}\), which uses 24 as a common denominator.
After finding the common denominator and combining the fractions, the resulting fraction represents the reciprocal of the unknown value. In the above example, the resulting fraction \( \frac{7}{24}\) is then equated to \( \frac{1}{f}\). To isolate \(f\), one must take the reciprocal of \( \frac{7}{24}\).
It is worth noting that to simplify the process, it is paramount to be comfortable with identifying least common denominators, combining fractions, and simplifying results. Mastery of these skills leads to a smoother and more accurate solution to problems that involve steps like these.
After finding the common denominator and combining the fractions, the resulting fraction represents the reciprocal of the unknown value. In the above example, the resulting fraction \( \frac{7}{24}\) is then equated to \( \frac{1}{f}\). To isolate \(f\), one must take the reciprocal of \( \frac{7}{24}\).
It is worth noting that to simplify the process, it is paramount to be comfortable with identifying least common denominators, combining fractions, and simplifying results. Mastery of these skills leads to a smoother and more accurate solution to problems that involve steps like these.
Reciprocal of Fractions
The reciprocal of a fraction is the inverted form of the original fraction where the numerator and the denominator are swapped. In algebraic terms, the reciprocal of \( \frac{a}{b}\) is \( \frac{b}{a}\), given that \(b \eq 0\). Working with reciprocals is a central part of solving equations that involve fractions, including the lens formula.
For example, when solving for the focal length \(f\) in the equation \( \frac{1}{p} + \frac{1}{q} = \frac{1}{f}\), after combining the fractions on the left side and equating them to \( \frac{1}{f}\), the next step is to find \(f\) itself by taking the reciprocal of the combined fraction. This means if the result is \( \frac{7}{24}\), then \(f = \frac{24}{7}\), which is an application of taking a reciprocal.
It's important to remember that the reciprocal of a fraction is distinctly different from simply inverting the sign of a fraction. The concept of a reciprocal becomes especially essential when dealing with various algebraic operations and is often used to solve equations or simplify expressions.
For example, when solving for the focal length \(f\) in the equation \( \frac{1}{p} + \frac{1}{q} = \frac{1}{f}\), after combining the fractions on the left side and equating them to \( \frac{1}{f}\), the next step is to find \(f\) itself by taking the reciprocal of the combined fraction. This means if the result is \( \frac{7}{24}\), then \(f = \frac{24}{7}\), which is an application of taking a reciprocal.
It's important to remember that the reciprocal of a fraction is distinctly different from simply inverting the sign of a fraction. The concept of a reciprocal becomes especially essential when dealing with various algebraic operations and is often used to solve equations or simplify expressions.
Lens Formula Application
The lens formula \( \frac{1}{p} + \frac{1}{q} = \frac{1}{f}\) is fundamental in optics and clearly demonstrates the relationship between the distance of an object from a lens \(p\), the distance of the image from the lens \(q\), and the lens's focal length \(f\). The formula is applied to find any one of these three values when the other two are known.
In practical applications, solving the lens formula often involves algebraic manipulation and the concept of reciprocals. For instance, in example (a), by substituting \(p\) and \(q\) into the lens formula and finding a common denominator, we were able to calculate the focal length \(f\). Conversely, if the focal length and either the object distance or image distance is known, the other can be determined, but one must be cautious of the physical limitations as seen in example (b) wherein an image distance equal to the focal distance doesn't yield a real-life situation.
The formula is quite versatile, as seen in example (c), where the focal length is given as a mixed fraction, requiring additional steps to convert it to an improper fraction before applying it to the lens formula. Such problems showcase the importance of understanding both basic fraction operations and the practical application of the lens formula to accurately determine distances related to a lens' focal point.
In practical applications, solving the lens formula often involves algebraic manipulation and the concept of reciprocals. For instance, in example (a), by substituting \(p\) and \(q\) into the lens formula and finding a common denominator, we were able to calculate the focal length \(f\). Conversely, if the focal length and either the object distance or image distance is known, the other can be determined, but one must be cautious of the physical limitations as seen in example (b) wherein an image distance equal to the focal distance doesn't yield a real-life situation.
The formula is quite versatile, as seen in example (c), where the focal length is given as a mixed fraction, requiring additional steps to convert it to an improper fraction before applying it to the lens formula. Such problems showcase the importance of understanding both basic fraction operations and the practical application of the lens formula to accurately determine distances related to a lens' focal point.
Other exercises in this chapter
Problem 99
Write the equation of the line that passes through the points (1,1) and (4,-2) . Express the equation in slope-intercept form.
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Simplify \(\left(x^{3} y^{2} z^{5}\right)^{6}\left(x^{2} y z\right)^{2}\)
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Reduce \(\frac{y^{2}-y-6}{y-3}\)
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Write \(6 a^{-3} b^{4} c^{-2} a^{-1} b^{-5} c^{3}\) so that only positive exponents appear.
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