Problem 117 - University Physics with Modern Physics

Question

(a) For a lens with focal length \(f,\) find the smallest distance possible between the object and its real image. (b) Graph the distance between the object and the real image as a function of the distance of the object from the lens. Does your graph agree with the result you found in part (a)?

Step-by-Step Solution

Answer

The smallest distance between the object and its real image is \(4f\). The graph verifies this result, showing a minimum at \(d_o = 2f\).

Step 1: Understand the Lens Formula

The lens formula is given by \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the distance of the object from the lens, and \( d_i \) is the distance of the image from the lens. Our goal is to find the minimum total distance \( d = d_o + d_i \).

Step 2: Express Total Distance and Minimize

Express the total distance \( d = d_o + d_i \). From the lens formula, we can write \( d_i = \frac{d_o f}{d_o - f} \). Then substituting \( d_i \) in \( d \), we get \( d = d_o + \frac{d_o f}{d_o - f} \). To find the minimum distance, we need the derivative of \( d \) with respect to \( d_o \) to be zero.

Step 3: Find the Derivative

Taking the derivative of the expression \( d = d_o + \frac{d_o f}{d_o - f} \), we use the quotient rule on \( \frac{d_o f}{d_o - f} \). This yields \( \frac{d}{dd_o} d = \frac{d_o - 2f}{(d_o - f)^2} \). Set this derivative equal to zero to find critical points: \( d_o - 2f = 0 \).

Step 4: Solve for Object Distance

From \( d_o - 2f = 0 \), solve for \( d_o \) to find \( d_o = 2f \). The corresponding image distance \( d_i \) is also \( 2f \), as \( d_i = \frac{d_o f}{d_o - f} = 2f \). Therefore, the minimum total distance \( d \) is \( 4f \).

Step 5: Graph the Distance Function

Plot the graph of \( d = d_o + d_i \) as a function of \( d_o \). This involves graphing \( d = d_o + \frac{d_o f}{d_o - f} \) for various values of \( d_o \) greater than \( f \). You will see the minimum point of the curve is at \( d_o = 2f \), confirming that the minimum distance \( d \) is \( 4f \).

Step 6: Conclusion from the Graph

The graph confirms that the minimum total distance between the object and its real image occurs when both \( d_o \) and \( d_i \) are equal to \( 2f \), making \( d = 4f \). This agrees with our earlier mathematical results.

Key Concepts

Focal Length

The focal length is a key property of a lens. It determines how strongly the lens converges or diverges light. When dealing with thin lenses, the focal length is the distance from the lens to the point where parallel rays of light converge (for a converging lens) or appear to diverge from (for a diverging lens). It is often represented by the symbol \( f \). Understanding its value is crucial as it impacts how images are formed.

• A positive focal length indicates a converging lens, which can focus incoming parallel light rays to a point.
• A negative focal length indicates a diverging lens, which spreads incoming parallel light rays.

In practical problems, knowing the focal length allows us to apply the lens formula to locate where the image will be formed relative to the lens. The lens formula: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] is fundamental in linking the focal length to distances that control image formation.

Real Image

A real image is formed when light rays converge after passing through a lens. Unlike a virtual image, a real image can be projected onto a screen. This type of image is always inverted compared to the object. For an image to be real:

• The object distance \( d_o \) should be greater than the focal length \( f \) of a converging lens.
• The image distance \( d_i \) is positive and on the opposite side of the lens, as per the convention for real images.

Real images are used in various applications, such as cameras and the human eye, where focusing light onto a surface is essential. In the exercise, finding the minimum distance between the object and its real image involves ensuring both are positioned optimally around the lens. By setting \( d_o = 2f \) and \( d_i = 2f \), we ensure a real image is produced at the minimum total distance \( 4f \).

Minimum Distance

Finding the minimum distance between the object and a real image in lens systems is a fascinating aspect of understanding optics. The challenge is to position the object such that the total distance \( d = d_o + d_i \) is minimized. To achieve this, we utilize the lens formula and calculus to determine: - The lens formula links the object distance \( d_o \), image distance \( d_i \), and the focal length \( f \).- By setting the derivative of the distance \( d \ = d_o + \frac{d_of}{d_o - f} \) with respect to \( d_o \) to zero, we find critical points.From the original problem, the critical configuration occurred at \( d_o = 2f \), resulting in \( d_i = 2f \), minimizing \( d \ = 4f \). This process highlights the balance in positioning that both optimizes space and maintains image clarity.

Object Distance

Object distance in the context of lenses refers to how far the object is located from the lens. It is symbolized by \( d_o \). Understanding its proper placement is essential for forming images using lenses.

• The object distance must always be greater than the focal length in a convergent lens to form a real image.
• Optimization problems often revolve around correctly setting the object distance to achieve specific criteria, like minimizing overall distance.

In our exercise, achieving the minimum total distance means setting the object distance to twice the focal length—\(d_o = 2f\). This ensures that both the object and resulting real image are precisely balanced around the lens. Correctly managing object distance allows control over image size and location, crucial in designing optical devices such as microscopes and projectors.