Problem 39 - Physics

Question

An object is located \(14.0 \mathrm{~cm}\) in front of a convex mirror, the image being \(7.00 \mathrm{~cm}\) behind the mirror. A second object, twice as tall as the first one, is placed in front of the mirror, but at a different location. The image of this second object has the same height as the other image. How far in front of the mirror is the second object located?

Step-by-Step Solution

Answer

The second object is 28 cm in front of the mirror.

Step 1: Understand Mirror Equation

The mirror equation is given by \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance. Conventionally for convex mirrors, the image distance \( d_i \) is negative.

Step 2: Calculate Focal Length

The first object's image distance \( d_{i1} = -7.0 \text{ cm} \) and object distance \( d_{o1} = 14.0 \text{ cm} \). Using the mirror equation, solve for \( f \): \( \frac{1}{f} = \frac{1}{14} - \frac{1}{7} = \frac{1}{14} + \frac{1}{-7} = \frac{1}{14} - \frac{2}{14} = -\frac{1}{14} \). Thus, \( f = -14.0 \text{ cm} \).

Step 3: Define Conditions for the Second Object

The second object's image height is the same as the first object's image height, meaning the magnification factors are the same for both objects. Since the second object is twice the height of the first but results in the same image height, its magnification in magnitude must be half the magnification of the first object.

Step 4: Find Magnifications

For the first object, the magnification \( m_1 = \frac{h_{i1}}{h_{o1}} = \frac{d_{i1}}{d_{o1}} = \frac{-7}{14} = -0.5 \). For the second object with the same image height as the first image \( h_{i2} = h_{i1} \), \( m_2 = \frac{h_{i2}}{h_{o2}} = \frac{h_{i1}}{2h_{o1}} = -0.25 \).

Step 5: Find Second Object Distance

Using the magnification \( m_2 = \frac{d_{i2}}{d_{o2}} = -0.25 \), we know \( d_{i2} \) is the same as \( d_{i1} = -7 \text{ cm} \) because both images have the same height. Substitute these values into the magnification formula: \( -0.25 = \frac{-7}{d_{o2}} \). Solving for \( d_{o2} \) gives \( d_{o2} = 28 \text{ cm} \).

Key Concepts

Mirror Equation

The mirror equation is an essential formula in optics to understand how mirrors create images. It relates the focal length of the mirror (\(f\)), the object distance from the mirror (\(d_o\)), and the image distance from the mirror (\(d_i\)). The equation is expressed as:
\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]For convex mirrors, which are the focus here, the image distance (\(d_i\)) is considered to be negative. This is due to the image being formed behind the mirror as opposed to on the same side as the object. This sign convention helps distinguish convex mirrors from concave mirrors where the image forms on the other side.
Understanding this equation is crucial as it serves as the foundation for calculating other important optics parameters like focal length and object distances.

Focal Length Calculation

The focal length of a mirror is a measure of how strongly it converges or diverges light. In the exercise, you are given the object distance \(d_o = 14.0 \text{ cm}\) and image distance \(d_i = -7.0 \text{ cm}\). Using the mirror equation:
\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]We substitute the given values:
\[\frac{1}{f} = \frac{1}{14} - \frac{1}{7}\ = -\frac{1}{14}\]Which yields the focal length \(f = -14.0 \text{ cm}\). The negative sign signifies that it is a convex mirror, where the focal point is behind the mirror. This calculation helps in identifying not only the type of mirror but also its power in terms of focusing abilities.

Magnification in Optics

Magnification is an important concept in optics that tells us how much larger or smaller an image is compared to the object. It is given by the formula:
\[m = \frac{h_i}{h_o} = \frac{d_i}{d_o}\]Where \(h_i\) is the image height, \(h_o\) is the object height, and \(d_i/d_o\) is the image-to-object distance ratio. For the first object, the magnification is \(-0.5\), indicating that the image is half the size of the object and inverted. For the second object, the magnification is \(-0.25\), which implies the image remains same in scale as the first image, despite differing object heights. This shows how magnification helps in comparing images created by different object placements in relation to size and direction.

Object and Image Distance

Object and image distances are crucial parameters in understanding image formation in mirrors. The object distance (\(d_o\)) is where the object is relative to the mirror, and image distance (\(d_i\)) is where the image appears.
For convex mirrors, \(d_i\) is always negative, representing the location of the virtual image formed behind the mirror. In step-by-step problem-solving, these distances are calculated to investigate how the object shifts affect the resulting image.
Identifying these distances helps us to solve many optics problems by seeing how changing these distances affects the size, orientation, and position of the image.

Physics Problem Solving

Physics problems, especially in optics, often require a systematic approach with attention to detail. The steps involve understanding the essential equations, substituting known values accurately, and applying sign conventions correctly. This includes interpreting physical phenomena mathematically.
In solving mirror problems, it is crucial to:

• Identify the type of mirror and relevant conventions.
• Apply the mirror equation.
• Carefully substitute the known values.
• Solve the equations for the unknowns.

Such methodical approaches ensure that one can correctly interpret and solve complex optics problems, providing a broad understanding of light behavior as it interacts with different surfaces.