AP Physics 2: Concave and Convex Mirrors

Understanding concave and convex mirrors is essential for mastering geometric optics, a cornerstone of AP Physics 2. These curved mirrors are not just academic curiosities; they are fundamental components in telescopes, headlights, security systems, and even your bathroom mirror. By learning to predict how they form images, you build a powerful toolkit for analyzing light's behavior and solving complex optical problems.

Foundations of Curved Mirrors

All curved mirrors are shaped like a section of a sphere. The principal axis is an imaginary line that passes through the center of this sphere and the mirror's midpoint. The center of curvature (C) is the point at the center of the imaginary sphere from which the mirror is cut, and its distance from the mirror is the radius of curvature (R). The vertex (V) is the geometric center of the mirror's surface. Midway between the vertex and the center of curvature lies the focal point (F). The distance from the vertex to the focal point is the focal length (f).

A critical relationship is $f=R/2$. The focal length is positive for concave mirrors and negative for convex mirrors, a sign convention that is vital for calculations. Concave mirrors, also called converging mirrors, curve inward like a spoon's bowl. They can collect light to a point. Convex mirrors, or diverging mirrors, bulge outward. They spread light apart, making them ideal for wide-angle views.

Image Formation with Concave Mirrors

The type of image formed by a concave mirror depends entirely on the object's location relative to the focal point (F) and center of curvature (C). You can determine image characteristics using two methods: precise ray tracing or the mirror equation. Ray tracing uses three principal rays:

1. A ray parallel to the principal axis reflects through the focal point.
2. A ray through the focal point reflects parallel to the principal axis.
3. A ray through the center of curvature reflects back on itself.

By drawing any two of these rays from the top of an object, you can find where they intersect (or appear to diverge from) to locate the image.

Let's analyze key object positions:

• Object beyond C: The image is real, inverted, reduced, and located between C and F.
• Object at C: The image is real, inverted, same size, and also at C.
• Object between C and F: The image is real, inverted, magnified, and located beyond C.
• Object at F: No image is formed (reflected rays are parallel).
• Object between F and the mirror: The image is virtual, upright, magnified, and appears behind the mirror. This is the principle behind shaving and makeup mirrors.

A real image is formed where light rays actually converge. It can be projected onto a screen. A virtual image is formed where light rays only appear to diverge from; it cannot be projected.

The Mirror Equation and Problem-Solving

The mirror equation provides a quantitative method to complement ray diagrams. It is expressed as: $$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$$ Here, $f$ is the focal length, $d_o$ is the object distance (always positive), and $d_i$ is the image distance. The sign convention is key:

• Focal Length ($f$): Positive for concave mirrors, negative for convex mirrors.
• Image Distance ($d_i$): Positive if the image is in front of the mirror (real), negative if behind (virtual).
• Object Distance ($d_o$): Always positive for real objects.

Example Problem: A 4.0 cm tall object is placed 30.0 cm in front of a concave mirror with a focal length of 10.0 cm. Find the image location and size.

Step 1: Identify knowns and the sign of f.
$d_o=+30.0$ cm, $h_o=+4.0$ cm, $f=+10.0$ cm (concave = positive).

Step 2: Apply the mirror equation to find $d_i$. $$\frac{1}{10.0}=\frac{1}{30.0}+\frac{1}{d_i}$$ $$\frac{1}{d_i}=\frac{1}{10.0}-\frac{1}{30.0}=\frac{3}{30.0}-\frac{1}{30.0}=\frac{2}{30.0}=\frac{1}{15.0}$$ $$d_i=+15.0\text{ cm}$$ The positive sign confirms a real image located 15.0 cm in front of the mirror.

Image Formation with Convex Mirrors

Convex mirrors have a negative focal length ($f<0$). For any real object placed in front of a convex mirror, the image is always virtual, upright, and reduced in size. It always appears behind the mirror, between the vertex and the focal point. This wide field of view is why convex mirrors are used for security and side-view mirrors (with the warning "Objects in mirror are closer than they appear").

Example Problem: A 3.0 cm tall object is placed 20.0 cm in front of a convex mirror with a focal length of -8.0 cm. Find the image location.

Step 1: Apply the mirror equation. $d_o=+20.0$ cm, $f=-8.0$ cm. $$\frac{1}{-8.0}=\frac{1}{20.0}+\frac{1}{d_i}$$ $$\frac{1}{d_i}=\frac{1}{-8.0}-\frac{1}{20.0}=-\frac{5}{40.0}-\frac{2}{40.0}=-\frac{7}{40.0}$$ $$d_i\approx -5.71\text{ cm}$$ The negative sign confirms a virtual image located 5.71 cm behind the mirror.

Magnification and Image Size

The magnification (m) tells you the image's orientation and size relative to the object. It is given by two equivalent formulas: $$m=\frac{h_i}{h_o}=-\frac{d_i}{d_o}$$ Here, $h_i$ and $h_o$ are the image and object heights, respectively. Sign conventions apply here too:

• Magnification Sign: Positive $m$ means an upright image. Negative $m$ means an inverted image.
• Magnification Magnitude: If $|m|>1$, the image is magnified. If $|m|<1$, the image is reduced. If $|m|=1$, the image is the same size.

Let's complete the first example with the concave mirror. We found $d_i=+15.0$ cm and $d_o=30.0$ cm. $$m=-\frac{d_i}{d_o}=-\frac{15.0}{30.0}=-0.50$$ The magnification is -0.50. The negative sign means the image is inverted. The magnitude of 0.50 means it is reduced to half the object's height. The image height is $h_i=m\times h_o=(-0.50)\times (4.0\text{ cm})=-2.0\text{ cm}$.

For the convex mirror example, $d_i=-5.71$ cm and $d_o=20.0$ cm. $$m=-\frac{(-5.71)}{20.0}=+0.285$$ The positive sign confirms an upright image, and the magnitude less than 1 confirms it is reduced.

Common Pitfalls

1. Sign Convention Errors: The most frequent mistake is using incorrect signs for $f$ and $d_i$. Always remember: concave $f$ is positive, convex $f$ is negative. A positive $d_i$ means a real image in front; a negative $d_i$ means a virtual image behind. Memorize and apply this consistently.

1. Misinterpreting "Behind the Mirror": For virtual images, $d_i$ is negative. Physically, no light is behind the mirror, but that's where the rays appear to originate. When a problem asks for the "image location," a final answer of $-12$ cm is perfectly correct and indicates a virtual image 12 cm behind the mirror's surface.

1. Confusing Magnification for Size: Magnification relates image size to object size. A magnification of 2 means the image is twice as tall, not that it is simply "larger." Furthermore, do not confuse the sign (orientation) with the magnitude (size). A magnification of $-0.3$ describes an inverted, reduced image.

1. Incorrect Ray Diagram Rays for Convex Mirrors: When drawing rays for a convex mirror, the ray parallel to the axis reflects as if it came from the focal point behind the mirror. The ray heading toward the focal point behind the mirror reflects parallel to the axis. Draw extensions behind the mirror with dashed lines to find the virtual image.

Summary

• The mirror equation, $\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$, and magnification formula, $m=-\frac{d_i}{d_o}$, are your primary quantitative tools for analyzing curved mirrors.
• Sign conventions are non-negotiable: Focal length ($f$) is positive for concave and negative for convex mirrors. Image distance ($d_i$) is positive for real images and negative for virtual images.
• Concave mirrors can produce real or virtual images depending on object position relative to F. Real images are inverted and projectable; virtual images are upright and magnified.
• Convex mirrors always produce virtual, upright, and reduced images for any real object position, giving them a wide field of view.
• Magnification ($m$) indicates orientation (sign) and relative size (magnitude). Use it with the height ratio $m=h_i/h_o$ to find unknown heights.
• Always sketch a quick ray diagram to check your mathematical solutions and build intuitive understanding of how image characteristics change with object position.