# Using Equations to Answer Mirror Questions

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• 0:02 What We Need to Know
• 1:01 Lens Equation
• 3:27 Example Calculation
• 5:48 Lesson Summary

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Lesson Transcript
Instructor: David Wood

David has taught Honors Physics, AP Physics, IB Physics and general science courses. He has a Masters in Education, and a Bachelors in Physics.

After watching this lesson, you will be able to answer questions on mirrors, stating whether an image is real or virtual, upright or inverted, and larger or smaller. A short quiz will follow.

## What We Need to Know

When you stand in front of a mirror, an image of you is created. You use this image to fix your hair, to put on make-up, or get an eyelash out of your eye. Those images are incredibly useful. And for a plane mirror, we all have a pretty good idea what that image will look like from experience.

But not all mirrors are flat. Convex and concave mirrors can produce images that are bigger than you, smaller than you, upright or inverted, and images that could either be real (images that form in front of the mirror so you can project onto a screen or blank piece of paper), or virtual (images that form behind the mirror so you can't project it onto a screen).

A regular, flat mirror creates an image that's the same size as you, virtual (behind the mirror), and upright. But for more complex mirror situations, we need to find ways to predict what the image will be like. One way we can do that is by carefully drawing ray diagrams with everything perfectly to scale. But a quicker and often more useful way is to use the power of math!

## Lens Equations

When it comes to mathematically figuring out what an image will look like, there are two main equations we need to use. The first is called the spherical mirror equation. It looks a lot like the thin lens equation, but for mirrors. Although, the terms are in slightly different places. It says that one over the object distance plus one over the focal length of the mirror is equal to one over the image distance. Just like with the thin lens equation, which is explored in more detail in another lesson, there are important sign conventions we need to know and use.

A positive focal length, f, represents a concave (or converging) mirror that brings light rays closer together. A negative focal length, -f, represents a convex (or diverging) mirror that spreads light rays further apart. The object distance is always negative because it's always behind the mirrored surface. But the image distance is positive if the image is real (formed in front of the mirror) and negative if the image is virtual (formed behind the mirror). These sign conventions are extremely similar to those used for the lens equation. The distances can be measured in meters or centimeters, whatever is convenient, as long as you use the same units for all your numbers.

By plugging numbers into the spherical mirror equation we can figure out where the image forms from left to right on our ray diagrams. We can, therefore, state whether an image is real or virtual. So, that's one of our three questions answered.

To figure out if the image is bigger or smaller and if it's upright or inverted, we need to use the magnification equation. This equation says that the magnification factor, M (for example, 3 for 3 times as big, or 0.5 for half as big), is equal to the image height, hi, divided by the object height, ho, which is equal to the image distance divided by the object distance. The image distance and the object distance follow the same sign convention as in the mirror equation. But now we also have a sign convention for the heights of the image. A positive image height represents an upright image, whereas a negative image height suggests an inverted image.

So, by using the magnification equation, we can figure out if the image is bigger or smaller by getting a number for the magnification. And, we can figure out if the image is upright or inverted by calculating the image height and looking at whether it has a positive or negative sign.

## Example Calculation

Such an equation heavy lesson definitely needs an example. Let's say you're stood in front of a spherical fun-house mirror, and your height is about 1.25 meters. The part of the mirror you're looking at is concave and has a focal length of 1.5 meters. If you stand at a distance of 2 meters away, is the image of you (a) real or virtual, (b) larger or smaller, and (c) upright or inverted?

Okay, so with any physics problem, the first thing we should do is write down what we know. We know that the focal length, f, of the mirror is 1.5 meters, and the object distance, do, is 2 meters. So, we want to know di to figure out if it's real or virtual, and we want to know M to see if it's larger or smaller. Last of all, we'll need to find hi to see if the image is upright or inverted.

So, let's start by using the mirror equation to calculate di. (1 / -2) + (1 / 1.5) = (1 / di). Just to be clear, it's -2 because with mirrors the object is always in front of the mirror, and it's positive 1.5 because it's a concave mirror. Plug numbers in, and we get 1/di = 1/1.5 + 1/-2. Take the reciprocal of both sides and solve.

This gives us an image distance of 6 meters. So, our image is 6 meters in front of the mirror - in front because it's a positive number. If the image forms in front of the mirror, if the number is positive, then it's a real image. So, that's the answer to part (a).

Next, we need to see if it's larger or smaller. The magnification, M, is equal to the image distance divided by the object distance. 6 / -2 gives us -3. So, the image is 3 times the size of you. Or, in other words, the image is larger. That's the answer to part (b).

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