13.2 Images Formed by Mirrors

Mirrors play a crucial role in optics, forming images through reflection. Different types of mirrors create different types of images based on their shape and the position of the object. Let's explore how these mirrors work and how they form images.

Focal Point of Concave Mirrors

Concave mirrors have a reflective surface that curves inward like the inside of a bowl. This unique shape affects how they handle light rays.

• When light rays parallel to the principal axis strike a concave mirror, they reflect and converge at a single point called the focal point
• This focal point is located in front of the mirror (on the same side as the incoming light)
• The distance from the mirror to this focal point is called the focal length (f)

Concave mirrors are also called converging mirrors because they bring light rays together. This convergence makes them useful in applications like telescopes, makeup mirrors, and car headlights where light concentration is needed.

Focal Point of Convex Mirrors

Convex mirrors curve outward like the outside of a sphere. This shape creates very different optical properties compared to concave mirrors.

• When parallel light rays strike a convex mirror, they reflect and diverge (spread out) as if they originated from a point behind the mirror
• This point is the virtual focal point of the convex mirror
• The focal point exists behind the mirror where light rays never actually go

Convex mirrors are diverging mirrors, spreading light out rather than concentrating it. This property gives them a wider field of view, making them ideal for security mirrors, side-view mirrors on vehicles, and other applications where a broader perspective is needed.

Focal Point of Plane Mirrors

Plane mirrors have a flat reflective surface with no curvature. Their optical properties are simpler than curved mirrors.

Plane mirrors don't converge or diverge light rays. When parallel light rays strike a plane mirror:

• They remain parallel after reflection
• The angle of incidence equals the angle of reflection for each ray
• There is no focal point at a finite distance (mathematically, the focal point is at infinity)

This is why plane mirrors produce images that are the same size as the object, unlike curved mirrors which can magnify or reduce image size.

Focal Point of Spherical Mirrors

Spherical mirrors (both concave and convex) have a surface that forms part of a sphere. Their focal points depend on their radius of curvature.

For spherical mirrors with small apertures (where light rays strike close to the principal axis):

• The focal length (f) is approximately half the radius of curvature (R): f = R/2
• This relationship applies to both concave and convex mirrors
• The focal length is positive for concave mirrors and negative for convex mirrors

This approximation works well for most practical applications but becomes less accurate for larger mirrors or rays far from the principal axis.

Real vs Virtual Images

Images formed by mirrors can be classified as either real or virtual, depending on how light rays interact.

Real images form when reflected light rays actually converge at a point in space:

• Light physically passes through each point of a real image
• Real images can be projected onto a screen
• Real images are typically inverted (upside-down)
• Only concave mirrors can form real images, and only when the object is beyond the focal point

Virtual images form when light rays appear to diverge from a point where they don't actually meet:

• Light doesn't physically pass through a virtual image
• Virtual images cannot be projected onto a screen
• Virtual images are typically upright
• Plane mirrors, convex mirrors, and concave mirrors (when the object is between the mirror and focal point) form virtual images

Image Location and Focal Length

The mirror equation relates the object distance (s₀), image distance (sᵢ), and focal length (f):

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

This equation works for all types of mirrors when using the proper sign conventions:

• For concave mirrors: f is positive
• For convex mirrors: f is negative
• Real images have positive sᵢ values
• Virtual images have negative sᵢ values

For plane mirrors, the focal length is infinite, so 1/f = 0. This simplifies the mirror equation to s₀ = -sᵢ, meaning the image appears exactly as far behind the mirror as the object is in front of it.

Image Magnification

The magnification (M) of an image tells us how much larger or smaller the image is compared to the object:

$M = \frac{h_i}{h_o} = -\frac{s_i}{s_o}$

Where:

• hᵢ is the image height
• h₀ is the object height
• sᵢ is the image distance
• s₀ is the object distance

The magnification formula provides two important pieces of information:

• The absolute value |M| tells us the size ratio between image and object
• The sign of M tells us the orientation: positive means upright, negative means inverted

For plane mirrors, M = 1, meaning the image is the same size as the object and upright.

Ray Diagrams for Mirrors

Ray diagrams are graphical tools that help us visualize where and how images form. They use specific light rays that are easy to trace:

1. Parallel Ray: A ray parallel to the principal axis reflects through (or appears to come from) the focal point
2. Focal Ray: A ray through (or aimed at) the focal point reflects parallel to the principal axis
3. Center Ray: A ray aimed at the center of curvature reflects back along the same path
4. Normal Ray: A ray striking the mirror perpendicularly to its surface at the vertex reflects back along the same path

The intersection of any two of these rays locates the image. For virtual images, we extend the reflected rays backward to find where they appear to intersect.

Ray diagrams help us determine:

• Whether the image is real or virtual
• Whether the image is upright or inverted
• The relative size of the image compared to the object
• The exact location of the image

Practice Problem 1: Concave Mirror Image Formation

Solution

First, let's use the mirror equation to find the image distance:

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

$\frac{1}{30 \text{ cm}} + \frac{1}{s_i} = \frac{1}{20 \text{ cm}}$

$\frac{1}{s_i} = \frac{1}{20 \text{ cm}} - \frac{1}{30 \text{ cm}} = \frac{3-2}{60 \text{ cm}} = \frac{1}{60 \text{ cm}}$

$s_i = 60 \text{ cm}$

The positive value of s₁ indicates that the image is real and located 60 cm in front of the mirror.

Now, let's calculate the magnification:

$M = -\frac{s_i}{s_o} = -\frac{60 \text{ cm}}{30 \text{ cm}} = -2$

The magnification is -2, which means:

• The image is 2 times larger than the object (30 cm tall)
• The negative sign indicates the image is inverted

Therefore, the image is:

• Located 60 cm in front of the mirror
• 30 cm tall
• Inverted (upside-down)
• Real (can be projected on a screen)

Practice Problem 2: Plane Mirror Image Location

Solution

For a plane mirror, the image distance equals the object distance but on the opposite side of the mirror. The total distance between the person and their image is twice the distance from the person to the mirror.

Initial situation:

• Person is 2 m from the mirror
• Image is 2 m behind the mirror
• Total distance between person and image = 4 m

Final situation:

• Total distance between person and image = 4 m - 1.5 m = 2.5 m
• If we call the final distance from person to mirror x, then:
  • 2x = 2.5 m
  • x = 1.25 m

Therefore, the person needs to walk 2 m - 1.25 m = 0.75 meters toward the mirror to reduce the distance between themselves and their image by 1.5 meters.