Problem 80 - Algebra and Trigonometry

Question

For a camera with a lens of fixed focal length \(F\) to focus on an object located a distance \(x\) from the lens, the film must be placed a distance \(y\) behind the lens, where \(F, x,\) and \(y\) are related by $$\frac{1}{x}+\frac{1}{y}=\frac{1}{F}$$ (See the figure.) Suppose the camera has a \(55-\mathrm{mm}\) lens \((F=55) .\) (a) Express \(y\) as a function of \(x\) and graph the function. (b) Express happens to the focusing distance \(y\) as the object moves far away from the lens? (c) What happens to the focusing distance \(y\) as the object moves close to the lens?

Step-by-Step Solution

Answer

(a) \(y = \frac{1}{(\frac{1}{55} - \frac{1}{x})}\), decreasing as \(x\) increases. (b) \(y\) approaches 55 as \(x\) goes to infinity. (c) \(y\) approaches infinity as \(x\) approaches 55.

Step 1: Solve for y

We start with the given formula: \(\frac{1}{x} + \frac{1}{y} = \frac{1}{F}\). Rearrange it to solve for \(y\). Subtract \(\frac{1}{x}\) from both sides: \(\frac{1}{y} = \frac{1}{F} - \frac{1}{x}\). To solve for \(y\), take the reciprocal of the entire expression: \[y = \frac{1}{\left(\frac{1}{F} - \frac{1}{x}\right)}.\] For a 55-mm lens where \(F = 55\), this becomes \[y = \frac{1}{\left(\frac{1}{55} - \frac{1}{x}\right)}.\]

Step 2: Graph the function y(x)

The function \(y = \frac{1}{\left(\frac{1}{55} - \frac{1}{x}\right)}\) is graphed using standard Cartesian coordinates. As \(x\) increases, \(y\) initially decreases until it approaches a minimum asymptotically. This is a hyperbolic relationship typically plotted over practically useful ranges of \(x\) that are greater than 55 since the object cannot be closer than the focal length in real-world scenarios.

Step 3: Analyze y(x) as x approaches infinity

As the object distance \(x\) increases to infinity, \(\frac{1}{x}\) approaches zero. Thus, the equation \(\frac{1}{y} = \frac{1}{55}\) simplifies, meaning \(y\) approaches \(55\). Therefore, the focusing distance \(y\) approaches the focal length \(F\) as the object moves very far away from the lens.

Step 4: Analyze y(x) as x approaches the focal length

As \(x\) approaches the focal length, which is \(x = 55\), \(\frac{1}{x}\) approaches \(\frac{1}{55}\). Consequently, \(\frac{1}{55} - \frac{1}{x}\) approaches zero, which would make \(y\) approach infinity. Therefore, the focusing distance \(y\) becomes extremely large as \(x\) approaches \(F\).

Key Concepts

Focal Length

The focal length, often denoted by the symbol \( F \), is a fundamental property of a lens. It represents the distance between the lens and the point where it brings light to focus. In the context of the camera, this distance is the measurement from the lens to the camera's sensor or film when focusing on distant objects.
The focal length determines the lens's magnification power and field of view.

• Longer focal lengths offer powerful magnification but a narrower field of view, ideal for distant subjects.
• Shorter focal lengths provide a wider field of view with less magnification, useful for capturing wider scenes.
  

In this exercise, the camera has a 55-mm focal length, which means it requires the film or sensor to be placed 55-mm away to focus on objects at a substantial distance.

Focusing Distance

Focusing distance refers to the distance from the lens to the film where the image comes to focus. This length changes based on the distance\( x \) of the object from the lens. The relationship between these distances is given by the lens formula:\[\frac{1}{x} + \frac{1}{y} = \frac{1}{F}\]Here, \( y \) is the focusing distance, which we solve to understand how it changes:\[y = \frac{1}{\left(\frac{1}{F} - \frac{1}{x}\right)}\]
Understanding how \( y \) changes as \( x \) varies provides insight into how lenses work:

• When the object moves further away (\( x \rightarrow \infty \)): \( y \) approaches the focal length \( F \).
• When the object is very close (\( x \rightarrow F \)): \( y \) approaches infinity, illustrating why a lens cannot focus on an object when it is too close.

Graphing Functions

Graphing the function \( y(x) = \frac{1}{\left(\frac{1}{55} - \frac{1}{x}\right)} \) helps visualize the relationship between the object distance \( x \) and the focusing distance \( y \).
In this exercise:

• The graph is hyperbolic, reflecting the reciprocal relationship.
• For practical purposes, \( x \) should be greater than 55, since the object must be farther from the lens than the focal length.
  

As you plot:

• You notice as \( x \) increases, \( y \) initially drops rapidly and levels off, approaching 55 mm, demonstrating the diminishing effect of distance on focus.
• This curve also reveals why close subjects need more focusing distance, showing \( y \) growing as \( x \) nears the focal length.