Problem 57 - Physics

Question

A converging lens \((f=25.0 \mathrm{~cm})\) is used to project an image of an object onto a screen. The object and the screen are \(125 \mathrm{~cm}\) apart, and between them the lens can be placed at either of two locations. Find the two object distances.

Step-by-Step Solution

Answer

The two object distances are approximately 90.45 cm and 34.55 cm.

Step 1: Understand the Lens Formula

The lens formula is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length of the lens, \( d_o \) is the object distance from the lens, and \( d_i \) is the image distance from the lens. In this problem, the total distance between the object and the screen is \( d_o + d_i = 125 \) cm.

Step 2: Set Up the Equations

Since \( d_o + d_i = 125 \) cm, we can express \( d_i \) in terms of \( d_o \) as \( d_i = 125 - d_o \). Substitute \( d_i = 125 - d_o \) into the lens formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{125 - d_o} \).

Step 3: Substitute Known Values

Substitute the given focal length \( f = 25 \) cm into the equation: \( \frac{1}{25} = \frac{1}{d_o} + \frac{1}{125 - d_o} \).

Step 4: Solve the Equation for \( d_o \)

Rearrange the equation \( \frac{1}{25} = \frac{1}{d_o} + \frac{1}{125 - d_o} \) to have a common denominator: \( \frac{(125 - d_o) + d_o}{d_o(125 - d_o)} = \frac{1}{25} \). Simplify to find: \( 125 = \frac{125d_o - d_o^2}{25} \).

Step 5: Solve the Quadratic Equation

Multiply through by 25 to clear the fraction: \( 3125 = 125d_o - d_o^2 \). Rearrange to the standard quadratic form: \( d_o^2 - 125d_o + 3125 = 0 \).

Step 6: Use the Quadratic Formula

Apply the quadratic formula: \( d_o = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -125 \), and \( c = 3125 \). Solve for \( d_o \): \( d_o = \frac{125 \pm \sqrt{125^2 - 4 \times 1 \times 3125}}{2} \).

Step 7: Calculate the Discriminant

Compute the discriminant: \( 125^2 - 4 \cdot 3125 = 15625 - 12500 = 3125 \). Calculate the square root: \( \sqrt{3125} = 55.9 \) (approximately).

Step 8: Find the Object Distances

Substitute back into the quadratic solution: \( d_o = \frac{125 \pm 55.9}{2} \). Calculate both possibilities: \( d_o \approx \frac{125 + 55.9}{2} = 90.45 \) cm and \( d_o \approx \frac{125 - 55.9}{2} = 34.55 \) cm.

Key Concepts

Converging Lens

A converging lens, often referred to as a convex lens, is specially shaped to magnify objects and focus light. This kind of lens is thicker at the center compared to its edges, which allows it to bend incoming parallel light rays toward a single point known as the focal point.

• Converging lenses are commonly used in devices like cameras, glasses, and microscopes.
• They help project clear images by directing light rays to converge.
• The distance from the center of the lens to the focal point is known as the focal length.

This particular lens property allows it to form real or virtual images depending on the object's position relative to the lens. Understanding how light interacts with converging lenses is crucial when working with optical devices.

Lens Formula

The lens formula is an essential equation in optics, allowing us to determine the relationship between the object distance, image distance, and focal length of a lens.
The formula is given as: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] where:

• \( f \) is the focal length of the lens.
• \( d_o \) is the distance from the object to the lens.
• \( d_i \) is the distance from the image to the lens.

The lens equation helps predict where an image will form based on the position of the object. By manipulating this formula, we can solve for unknown quantities if other parameters are given. It's a critical tool in understanding and designing optical systems.

Quadratic Equation

In optics, solving lens problems can lead to a quadratic equation. Quadratic equations are polynomial equations of the second degree, typically written in the form:\[ ax^2 + bx + c = 0 \] For the problem at hand, when the lens formula leads to nonlinear relationships, rearranging the equation to form a quadratic is necessary:\[ d_o^2 - 125d_o + 3125 = 0 \] This can be solved using the quadratic formula:\[ d_o = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where:

• \( a = 1 \)
• \( b = -125 \)
• \( c = 3125 \)

This formula yields possible values for the object distance. Solving these equations is integral for calculating precise optical measurements.

Focal Length

Focal length is a fundamental concept in optics, indicating the distance from the lens to its focal point where light rays converge. For a converging lens, the focal length determines how strongly it converges or focuses light.

• Positive focal length indicates a converging lens.
• A shorter focal length means the lens is stronger in converging light rays.
• It is measured in centimeters or meters depending on the context.

In this exercise, the focal length is given as 25 cm, crucial in applying the lens formula to find the object and image distances. By understanding the focal length, one can predict how a lens will behave in different setups, which is essential for designing lenses for specific purposes such as photography or magnification.