Problem 21 - Physics

Question

A small postage stamp is placed in front of a concave mirror (radius \(=R\) ), such that the image distance equals the object distance. (a) In terms of \(R\), what is the object distance? (b) What is the magnification of the mirror? (c) State whether the image is upright or inverted relative to the object. Draw a ray diagram to guide your thinking.

Step-by-Step Solution

Answer

(a) The object distance is \( R \). (b) The magnification is -1. (c) The image is inverted.

Step 1: Understanding the Mirror Equation

The mirror equation is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance. Since the image distance equals the object distance (\( d_i = d_o \)), we can simplify the equation for this specific setup.

Step 2: Find the Focal Length

Given that the radius of curvature \( R \), the focal length \( f \) is half of the radius, so \( f = \frac{R}{2} \).

Step 3: Substitute into the Mirror Equation

Substituting \( d_i = d_o \) and \( f = \frac{R}{2} \) into the mirror equation gives us \( \frac{1}{\frac{R}{2}} = \frac{1}{d_o} + \frac{1}{d_o} \). This simplifies to \( \frac{2}{R} = \frac{2}{d_o} \).

Step 4: Solve for the Object Distance

Solving \( \frac{2}{R} = \frac{2}{d_o} \), we find that \( d_o = R \).

Step 5: Calculate the Magnification

Magnification \( m \) is given by \( m = \frac{-d_i}{d_o} \). With \( d_i = d_o = R \), magnification becomes \( m = \frac{-R}{R} = -1 \).

Step 6: Determine Image Orientation

A magnification of \(-1\) indicates that the image is inverted relative to the object.

Key Concepts

Mirror Equation

The mirror equation is vital when working with mirrors, especially concave ones. This equation is expressed as \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where:

• \( f \) represents the focal length of the mirror,
• \( d_o \) is the object distance (distance between the object and the mirror),
• \( d_i \) is the image distance (distance between the image and the mirror).

This formula helps in finding out how an image is formed by the mirror. For this specific scenario where the image distance equals the object distance (\( d_i = d_o \)), the equation simplifies significantly. This symmetry allows us to deduce other properties of the image, such as its size and orientation.

Focal Length

The focal length is an intrinsic property of concave mirrors. It is the distance from the mirror to the focal point, where parallel rays of light either converge or appear to converge after reflecting off the mirror. The focal length \( f \) is directly related to the radius of curvature \( R \) of the mirror:

• The focal length is half the radius of curvature, or \( f = \frac{R}{2} \).

In practice, this means that the closer the curvature of the mirror, the shorter the focal length. A short focal length indicates a highly curved mirror. Understanding the focal length is necessary when applying the mirror equation, as it directly influences the shape and location of the image produced.

Magnification

Magnification in optics measures how much larger or smaller the image is compared to the object. For mirrors, magnification (\( m \)) is calculated as \( m = \frac{-d_i}{d_o} \).

• It reveals the ratio of the image height to the object height.
• A negative magnification signifies that the image is inverted.

In our exercise, with \( d_i = d_o = R \), the magnification becomes \( m = \frac{-R}{R} = -1 \). This means the image is the same size as the object but flipped upside down. The negative sign alerts us to the inverted nature of the image, which is typical for images formed by concave mirrors when the object is placed outside the focal length.

Inverted Image

In optics, an inverted image is one that is upside down compared to the object. This can occur in concave mirrors under certain conditions. When the magnification \( m \) is negative, it signifies inversion.

• This happens when the object is placed beyond the focal point of a concave mirror.
• An inverted image usually suggests a real image, as it may be projected onto a screen.

In the given problem, since the magnification is \(-1\), the image is confirmed to be inverted. No alterations to the object's size occur; only the orientation changes. Understanding the concept of an inverted image helps predict how the image will look in practical applications, which is crucial for designing optical devices.