Mirror Equation: Videos & Practice Problems

Spherical mirrors, including concave and convex types, are defined by their radius of curvature (r) and focal length (f = r/2). The mirror equation, $(\frac{1}{s}+\frac{1}{s}=\frac{1}{f}$, helps determine image location and characteristics. Concave mirrors produce real, inverted images, while convex mirrors yield virtual, upright images. Magnification indicates image size relative to the object, with positive values indicating upright images and negative values indicating inverted images.

Mirror Equation

Understanding the behavior of images produced by mirrors is essential in optics, particularly when dealing with spherical mirrors, which are the most common type encountered. A spherical mirror is essentially a segment of a sphere, characterized by its radius of curvature, denoted as \( r \). The focal length \( f \) of a spherical mirror is directly related to this radius and is calculated using the formula:

\[ f = \frac{r}{2} \]

To determine the image location, we utilize the mirror equation, which is expressed as:

\[ \frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f} \]

Here, \( d_o \) represents the object distance (the distance from the object to the mirror), and \( d_i \) is the image distance (the distance from the image to the mirror). The focal length \( f \) is positive for concave mirrors and negative for convex mirrors, following specific sign conventions. Concave mirrors, which converge light, have a positive focal length and can produce both real and virtual images. In contrast, convex mirrors diverge light, possess a negative focal length, and can only create virtual images.

Real images, formed by concave mirrors, are characterized by a positive image distance and are always inverted. Conversely, virtual images, which can be produced by both concave and convex mirrors, have a negative image distance and are upright. For plane mirrors, the focal length is considered infinite, and they can only produce virtual images.

The magnification \( m \) of an image, which indicates its size relative to the object, is given by the equation:

\[ m = -\frac{d_i}{d_o} \]

A positive magnification indicates an upright image, while a negative magnification signifies an inverted image. The absolute value of the magnification also reveals the relative size of the image compared to the object. For instance, a magnification of 2 means the image is twice as tall as the object, while a magnification of 0.5 indicates the image is half as tall.

To illustrate these concepts, consider a scenario where a 1.4-meter tall person stands 1 meter in front of a plane mirror. Since the focal length is infinite, the mirror equation simplifies to show that the image distance is negative 1 meter, confirming that the image is virtual. Using the magnification equation, we find:

\[ m = -\frac{-1 \text{ m}}{1 \text{ m}} = 1 \]

This positive magnification indicates that the image is upright and, since the magnification is 1, the height of the image equals the height of the object, which is 1.4 meters.

In summary, the mirror equation and magnification formula provide a comprehensive framework for analyzing the properties of images formed by different types of mirrors, enhancing our understanding of optical phenomena.

Example 1

To determine the characteristics of an image formed by a convex mirror, we can use the mirror equation and the concept of magnification. In this example, a 5-centimeter tall object is placed 10 centimeters in front of a convex mirror with a radius of curvature of 2 centimeters.

First, we need to find the focal length (f) of the mirror. The focal length is calculated using the formula:

$$ f = -\frac{r}{2} $$

where \( r \) is the radius of curvature. For a convex mirror, the focal length is negative. Thus, with \( r = 2 \) cm, we have:

$$ f = -\frac{2}{2} = -1 \text{ cm} $$

Next, we apply the mirror equation:

$$ \frac{1}{f} = \frac{1}{s_o} + \frac{1}{s_i} $$

Here, \( s_o \) is the object distance (10 cm) and \( s_i \) is the image distance. Rearranging the equation to solve for \( s_i \) gives us:

$$ \frac{1}{s_i} = \frac{1}{f} - \frac{1}{s_o} $$

Substituting the known values:

$$ \frac{1}{s_i} = \frac{1}{-1} - \frac{1}{10} $$

Finding a common denominator, we have:

$$ \frac{1}{s_i} = -\frac{10}{10} - \frac{1}{10} = -\frac{11}{10} $$

Taking the reciprocal to find \( s_i \):

$$ s_i = -\frac{10}{11} \approx -0.91 \text{ cm} $$

The negative sign indicates that the image is virtual, meaning it is located behind the mirror. Since virtual images formed by convex mirrors are always upright, we conclude that the image is upright.

To find the height of the image, we use the magnification formula:

$$ m = -\frac{s_i}{s_o} $$

Substituting the values:

$$ m = -\left(-\frac{10}{11}\right) \cdot \frac{1}{10} = \frac{1}{11} $$

The height of the image (\( h_i \)) can then be calculated as:

$$ h_i = m \cdot h_o $$

where \( h_o \) is the height of the object (5 cm):

$$ h_i = \frac{1}{11} \cdot 5 \approx 0.45 \text{ cm} $$

In summary, the image characteristics are as follows: the image is located approximately 0.91 centimeters behind the mirror, it is virtual and upright, and its height is approximately 0.45 centimeters.

A 4 cm tall object is placed 15 cm in front of a concave mirror with a focal length of 5 cm. Where is the image produced? Is this image real or virtual? Is it upright or inverted? What is the height of the image?

• A. s_i = 7.5cm; Real; Inverted, 2cm
• B. s_i = 15cm; Real; Inverted, 4cm
• C. s_i = 0.13 cm; Real; Inverted, 0.0087cm
• D. s_i = -7.5cm; Real; Inverted, -2cm

Answer: (A)

You want to produce a mirror that can produce an upright image that would be twice as tall as the object when placed 5 cm in front of it. What shape should this mirror be? What radius of curvature should the mirror have?

• A. concave; 20cm
• B. convex; 20cm
• C. concave; 10cm
• D. convex; 10 cm
• E. It is not possible

Answer: (A)

Here’s what students ask on this topic:

What is the mirror equation and how is it used?

The mirror equation is a fundamental formula used to determine the location and characteristics of images formed by spherical mirrors. It is given by:

$\frac{1}{\mathrm{s_o}}+\frac{1}{\mathrm{s_i}}=\frac{1}{f}$

where $\mathrm{s_o}$ is the object distance, $\mathrm{s_i}$ is the image distance, and $f$ is the focal length of the mirror. This equation helps determine whether the image is real or virtual, upright or inverted, and its location relative to the mirror.

How do you determine the focal length of a spherical mirror?

The focal length of a spherical mirror is determined by its radius of curvature. For any spherical mirror, the focal length $f$ is given by:

$f=\frac{r}{2}$

where $r$ is the radius of curvature of the mirror. This means the focal length is half the radius of curvature.

What are the sign conventions for concave and convex mirrors?

In the context of the mirror equation, concave and convex mirrors have specific sign conventions:

• Concave mirrors (converging mirrors) have a positive focal length. They can produce both real and virtual images, with real images having a positive image distance and virtual images having a negative image distance.
• Convex mirrors (diverging mirrors) have a negative focal length. They can only produce virtual images, which always have a negative image distance.

How do you calculate the magnification of an image formed by a mirror?

The magnification $m$ of an image formed by a mirror is given by the equation:

$m=\; -\frac{\mathrm{s_i}}{\mathrm{s_o}}$

where $\mathrm{s_i}$ is the image distance and $\mathrm{s_o}$ is the object distance. The negative sign indicates the orientation of the image. A positive magnification means the image is upright, while a negative magnification means the image is inverted.

What types of images can concave and convex mirrors produce?

Concave mirrors can produce both real and virtual images. Real images are formed when light converges and are always inverted with a positive image distance. Virtual images are formed when light appears to diverge from a point behind the mirror and are always upright with a negative image distance.

Convex mirrors, on the other hand, can only produce virtual images. These images are always upright and have a negative image distance because the light appears to diverge from a point behind the mirror.