Students will learn how light reflects off mirrors. Flat mirrors, concave mirrors and convex mirrors are all covered.
Key Equations
\begin{align*} \frac{1}{f} = \frac{1}{d_o} +\frac{1}{d_i} \; \end{align*}; The len's maker's equation
Where f is the focal length of the mirror, \begin{align*} d_o \end{align*} is the distance of the object from the mirror and \begin{align*} d_i \end{align*} is the distance the image is formed from the mirror.
\begin{align*} M = \frac{-d_i}{d_0} \end{align*}
The size of an object’s image is larger (or smaller) than the object itself by its magnification, \begin{align*}M\end{align*}. The level of magnification is proportional to the ratio of \begin{align*}d_i\end{align*} and \begin{align*}d_o\end{align*}. An image that is double the size of the object would have magnification \begin{align*}M = 2\end{align*}.
\begin{align*} R = 2f \end{align*}
The radius of curvature of a mirror is twice its focal length
- Mirrors are made from highly reflective metal that is applied to a curved or flat piece of glass. Converging mirrors can be used to focus light – headlights, telescopes, satellite TV receivers, and solar cookers all rely on this principle.
- Converging mirrors (also known as concave mirrors) are curved towards the incoming light and focus parallel rays at the focal point. Diverging mirrors (also known as convex mirrors) are curved away from the incoming light and do not focus, but scatter the light instead.
- The focal length, \begin{align*}f\end{align*}, of a lens or mirror is the distance from the surface of the lens or mirror to the place where the light is focused. This is called the focal point or focus. For diverging mirrors, there is no true focal point (i.e. real light does not focus), so the focal length is negative.
- When light rays converge in front of a mirror, a real image is formed. Real images are useful in that you can place photographic film at the physical location of the real image, expose the film to the light, and make a two-dimensional representation of the world, a photograph.
- When light rays diverge in front of a mirror, a virtual image is formed. A virtual image is formed by your brain tracing diverging rays backwards and is kind of a trick, like the person you see “behind” a mirror’s surface when you brush your teeth (there's obviously no real light focused behind a mirror!). Since virtual images aren’t actually “anywhere,” you can’t place photographic film anywhere to capture them.
- Real images are upside-down, or inverted. You can make a real image of an object by putting it farther from a mirror or lens than the focal length. Virtual images are typically right-side-up. You can make virtual images by moving the mirror or lens closer to the object than the focal length.
In the problems below that consider converging or diverging mirrors, you will do a careful ray tracing with a ruler (including the extrapolation of rays for virtual images). It is best if you can use different colors for the three different ray tracings. When sketching diverging rays, you should use dotted lines for the extrapolated lines behind a mirror in order to produce the virtual image. When comparing measured distances and heights to calculated distances and heights, values within \begin{align*}10\end{align*}% are considered “good.” Use the Table (below) as your guide.
Mirror type | Ray tracings |
---|---|
Converging mirrors (concave) |
Ray #1: Leaves tip of candle, travels parallel to optic axis, reflects back through focus. Ray #2: Leaves tip, travels through focus, reflects back parallel to optic axis. Ray #3: Leaves tip, reflects off center of mirror with an angle of reflection equal to the angle of incidence. |
Diverging mirrors (convex) |
Ray #1: Leaves tip, travels parallel to optic axis, reflects OUTWARD by lining up with focus on the OPPOSITE side as the candle. Ray #2: Leaves tip, heads toward the focus on the OPPOSITE side, and emerges parallel to the optic axis. Ray #3: Leaves tip, heads straight for the mirror center, and reflects at an equal angle. |
Example 1
In the situation illustrated below, the object is set up .5 m away from a converging mirror. If the focal length of the lens is .2 m, determine (a) the location of the real image, and (b) the magnification of the image.
Solution
(a): In order to determine the location of the image, we'll use the lens maker's equation.
\begin{align*} \frac{1}{f}&=\frac{1}{d_o} + \frac{1}{d_i}\\ \frac{1}{d_i}&=\frac{1}{f} - \frac{1}{d_o}\\ \frac{1}{d_i}&=\frac{1}{.2\;\text{m}} - \frac{1}{.5\;\text{m}}\\ d_i&=.33\;\text{m}\\ \end{align*}
(b): Now that we have the location of the image, we can find the magnification.
\begin{align*} M&=\frac{-d_i}{d_o}\\ M&=\frac{-.33\;\text{m}}{.5\;\text{m}}\\ M&=-\frac{2}{3}\\ \end{align*}
Watch this Explanation
Time for Practice
- Here’s an example of the “flat mirror problem.” Marjan is looking at herself in the mirror. Assume that her eyes are 10 cm below the top of her head, and that she stands \begin{align*}180 \;\mathrm{cm}\end{align*} tall. Calculate the minimum length flat mirror that Marjan would need to see her body from eye level all the way down to her feet. Sketch at least \begin{align*}3\end{align*} ray traces from her eyes showing the topmost, bottommost, and middle rays.
- Consider a concave mirror with a focal length equal to two units, as shown below.
- Carefully trace three rays coming off the top of the object in order to form the image.
- Measure \begin{align*}d_o\end{align*} and \begin{align*}d_i\end{align*}.
- Use the mirror/lens equation to calculate \begin{align*}d_i\end{align*}.
- Find the percent difference between your measured \begin{align*}d_i\end{align*} and your calculated \begin{align*}d_i\end{align*}.
- Measure the magnification \begin{align*}M\end{align*} and compare it to the calculated magnification.
- Consider a concave mirror with unknown focal length that produces a virtual image six units behind the mirror.
- Calculate the focal length of the mirror and draw an \begin{align*}\times\end{align*} at the position of the focus.
- Carefully trace three rays coming off the top of the object and show how they converge to form the image.
- Does your image appear bigger or smaller than the object? Calculate the expected magnification and compare it to your sketch.
- Consider a convex mirror with a focal length equal to two units.
- Carefully trace three rays coming off the top of the object and form the image.
- Measure \begin{align*}d_o\end{align*} and \begin{align*}d_i\end{align*}.
- Use the mirror/lens equation to calculate \begin{align*}d_i\end{align*}.
- Find the percent difference between your measured \begin{align*}d_i\end{align*} and your calculated \begin{align*}d_i\end{align*}.
- Measure the magnification \begin{align*}M\end{align*} and compare it to the calculated magnification.
- Above is a diagram showing how to make a “ghost light bulb.” The real light bulb is below the box and it forms an image of the exact same size right above it. The image looks very real until you try to touch it. What is the focal length of the concave mirror?
Answers to Selected Problems
- \begin{align*}85 \;\mathrm{cm}\end{align*}
- C. \begin{align*}+4\end{align*} units e. \begin{align*}M=-1\end{align*}
- a. \begin{align*}6\end{align*} units c. bigger; \begin{align*}M = 3\end{align*}
- c. \begin{align*}1.5\end{align*} units e. \begin{align*}M=2/3\end{align*}
- \begin{align*}32\;\mathrm{cm}\end{align*}