Problem 114 - Physics

Question

An object is in front of a converging lens \((f=0.30 \mathrm{~m})\). The magnification of the lens is \(m=4.0 .\) (a) Relative to the lens, in what direction should the object be moved so that the magnification changes to \(m=-4.0 ?\) (b) Through what distance should the object be moved?

Step-by-Step Solution

Answer

(a) Move the object away from the lens. (b) Move it by 0.15 m.

Step 1: Understanding Magnification Formula

The magnification of a lens can be related to the object distance (\(d_o\)) and the image distance (\(d_i\)) with the equation \(m = -\frac{d_i}{d_o}\). Initially, \(m = 4.0\), therefore \(-\frac{d_i}{d_o} = 4\). This implies \(d_i = -4d_o\).

Step 2: Using Lens Formula

Using the lens formula \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), with \(f = 0.30\, \text{m}\), and substituting \(d_i = -4d_o\), gives \(\frac{1}{0.3} = \frac{1}{d_o} - \frac{1}{4d_o}\).

Step 3: Simplifying the Lens Formula

Simplify the expression: \(\frac{1}{0.3} = \frac{3}{4d_o}\). Solving for \(d_o\), we find \(d_o = \frac{3}{4} \times 0.3 = 0.225\, \text{m}\).

Step 4: Calculate New Object Distance for New Magnification

For \(m = -4.0\), \(-\frac{d_i}{d_o} = -4\) means \(d_i = 4d_o\). Using the lens formula \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) and substituting \(d_i = 4d_o\), we have \(\frac{1}{0.3} = \frac{1}{d_o} + \frac{1}{4d_o}\).

Step 5: Solving for New Object Distance

Solving \(\frac{1}{0.3} = \frac{5}{4d_o}\), we find \(d_o = \frac{5}{4} \times 0.3 = 0.375\, \text{m}\).

Step 6: Finding the Direction and Distance to Move the Object

Since \(d_o\) increases from \(0.225\, \text{m}\) to \(0.375\, \text{m}\), the object must be moved away from the lens by a distance \(0.375\, \text{m} - 0.225\, \text{m} = 0.15\, \text{m}\).

Key Concepts

magnification

Magnification in optics is a measure of how much larger or smaller an image appears compared to the object itself. The magnification factor, represented as \( m \), is defined by the ratio \( m = \frac{h_i}{h_o} \), where \( h_i \) is the image height and \( h_o \) the object height. However, in the context of lenses, magnification can also be expressed in terms of the distances of the object and image from the lens using the formula \( m = -\frac{d_i}{d_o} \). This formula helps us understand two things:

• The negative sign indicates that when the image distance \( d_i \) and object distance \( d_o \) are in the same half-plane relative to the lens, the image is inverted.
• The magnitude of \( m \) tells us how many times the image is scaled compared to the object.

In this exercise, the magnification changes from a positive to a negative value, indicating a switch from an upright to an inverted image.

lens formula

The lens formula is a basic equation used in optics to determine the relationship between the focal length \( f \), the object distance \( d_o \), and the image distance \( d_i \). The formula is given by:\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]This formula is pivotal in calculating the position and nature (real or virtual) of the images formed by lenses. Understanding this formula requires an understanding that:

• \( f \) is the fixed property of a lens and tells us how strongly it can converge or diverge light.
• The object and image distances, \( d_o \) and \( d_i \), change depending on the position of the object in relation to the lens.

In the given problem, utilizing \( d_i = -4d_o \) for the initial condition and \( d_i = 4d_o \) for the desired change helps in applying this formula to find the required object movements.

object distance

The object distance, \( d_o \), is the distance from the object to the lens. It is a crucial factor in determining the characteristics of the image formed by the lens. By altering the object's position relative to the lens, you can change both the size and orientation of the image it forms.This exercise requires calculating \( d_o \) for different magnifications. Initially, for a magnification of 4, \( d_o \) is calculated as 0.225 meters using the lens and magnification formulas. When the magnification changes to -4, indicating an inverted image, \( d_o \) needs to be adjusted to 0.375 meters. This adjustment emphasizes the crucial role of object distance in manipulating image properties in optics.

image distance

Image distance, \( d_i \), refers to the distance from the image to the lens. Just like object distance, it significantly influences the nature of the image produced by a lens—whether it's real or virtual, upright or inverted.In this exercise, the initial scenario with \( m = 4 \) gives us \( d_i = -4d_o \), resulting in an image distance of -0.9 meters when \( d_o \) is 0.225 meters. For the scenario where the magnification shifts to \( m = -4 \), \( d_i \) changes to 4 times \( d_o \), or 1.5 meters when the object distance is corrected to 0.375 meters. This approach highlights how changes in image distance directly influence the characteristics of the light path through the lens, manipulating the perceived size and orientation of the resultant image.