# Angle of incidence and reflection relationship questions

### BBC Bitesize - Higher Physics - Refraction of light - Revision 3

A tutorial including problems with solutions on reflection of light rays. of incidence (2) The angle of incidence and the angle of reflection have the same size. Learn about and revise reflection and refraction of waves with GCSE Bitesize Physics. angle of incidence = angle of reflection. For example, if a light ray hits a. What relationship holds between the angle of incidence and the angle of reflection? The relationship between the incident angle and the refraction angle is that.

This peculiarity about the diamond-air boundary plays an important role in the brilliance of a diamond gemstone. Having a small critical angle, light has the tendency to become "trapped" inside of a diamond once it enters.

### The Critical Angle

A light ray will typically undergo TIR several times before finally refracting out of the diamond. Because the diamond-air boundary has such a small critical angle due to diamond's large index of refractionmost rays approach the diamond at angles of incidence greater than the critical angle. This gives diamond a tendency to sparkle. The effect can be enhanced by the cutting of a diamond gemstone with a strategically planned shape.

The diagram below depicts the total internal reflection within a diamond gemstone with a strategic and a non-strategic cut. Use the Find the Critical Angle widget below to investigate the effect of the indices of refraction upon the critical angle. Simply enter the index of refraction values; then click the Calculate button to view the result. Use the widget as a practice tool.

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Check Your Understanding 1. Suppose that the angle of incidence of a laser beam in water and heading towards air is adjusted to degrees. Use Snell's law to calculate the angle of refraction? Explain your result or lack of result.

See Answer Good luck! This problem has no solution. The angle of incidence is greater than the critical angle, so TIR occurs. There is no angle of refraction. Aaron Agin is trying to determine the critical angle of the diamond-glass surface. He looks up the index of refraction values of diamond 2.

### The reflection and refraction of light

Aaron hits the calculator and throws it own the ground a few times; he then repeats the calculation with the same result. He then utters something strange about the pizza he had slopped on it the evening before and runs out of the library with a disappointed disposition. What is Aaron's problem?

That is, what is the problem with his method of calculating the critical angle? See Answer Poor Aaron! It's not your pizza that's causing the problem; its your inappropriate use of the equation.

The chief and parallel rays meet at the tip of the image. The third ray, the focal ray, is a mirror image of the parallel ray. The focal ray is drawn from the tip of the object through or towards the focal point, reflecting off the mirror parallel to the principal axis. All three rays should meet at the same point. A ray diagram for a concave mirror is shown above. This shows a few different things. For this object, located beyond the center of curvature from the mirror, the image lies between the focal point F and the center of curvature.

The image is inverted compared to the object, and it is also a real image, because the light rays actually pass through the point where the image is located. With a concave mirror, any object beyond C will always have an image that is real, inverted compared to the object, and between F and C.

You can always trade the object and image places that just reverses all the arrows on the ray diagramso any object placed between F and C will have an image that is real, inverted, and beyond C. What happens when the object is between F and the mirror?

You should draw the ray diagram to convince yourself that the image will be behind the mirror, making it a virtual image, and it will be upright compared to the object.

A ray diagram for a convex mirror What happens with a convex mirror? In this case the ray diagram looks like this: As the ray diagram shows, the image for a convex mirror is virtual, and upright compared to the object. A convex mirror is the kind of mirror used for security in stores, and is also the kind of mirror used on the passenger side of many cars "Objects in mirror are closer than they appear.

A convex mirror will reflect a set of parallel rays in all directions; conversely, it will also take light from all directions and reflect it in one direction, which is exactly how it's used in stores and cars. The mirror equation Drawing a ray diagram is a great way to get a rough idea of how big the image of an object is, and where the image is located.

We can also calculate these things precisely, using something known as the mirror equation. The textbook does a nice job of deriving this equation in section Magnification In most cases the height of the image differs from the height of the object, meaning that the mirror has done some magnifying or reducing. The magnification, m, is defined as the ratio of the image height to the object height, which is closely related to the ratio of the image distance to the object distance: A magnification of 1 plus or minus means that the image is the same size as the object.

If m has a magnitude greater than 1 the image is larger than the object, and an m with a magnitude less than 1 means the image is smaller than the object. If the magnification is positive, the image is upright compared to the object; if m is negative, the image is inverted compared to the object.

Sign conventions What does a positive or negative image height or image distance mean? To figure out what the signs mean, take the side of the mirror where the object is to be the positive side. Any distances measured on that side are positive. Distances measured on the other side are negative. When the image distance is positive, the image is on the same side of the mirror as the object, and it is real and inverted.

When the image distance is negative, the image is behind the mirror, so the image is virtual and upright. A negative m means that the image is inverted. Positive means an upright image.

Steps for analyzing mirror problems There are basically three steps to follow to analyze any mirror problem, which generally means determining where the image of an object is located, and determining what kind of image it is real or virtual, upright or inverted.

Step 1 - Draw a ray diagram. The more careful you are in constructing this, the better idea you'll have of where the image is. Step 2 - Apply the mirror equation to determine the image distance. Or to find the object distance, or the focal length, depending on what is given. Step 3 - Make sure steps 1 and 2 are consistent with each other.

An example A Star Wars action figure, 8. Where is the image? How tall is the image?

## The Law of Reflection

What are the characteristics of the image? The first step is to draw the ray diagram, which should tell you that the image is real, inverted, smaller than the object, and between the focal point and the center of curvature. The location of the image can be found from the mirror equation: The image distance is positive, meaning that it is on the same side of the mirror as the object.

This agrees with the ray diagram. Note that we don't need to worry about converting distances to meters; just make sure everything has the same units, and whatever unit goes into the equation is what comes out. Calculating the magnification gives: Solving for the image height gives: The negative sign for the magnification, and the image height, tells us that the image is inverted compared to the object.

Activity 2: Angle of Incidence v. Angle of Reflection

To summarize, the image is real, inverted, 6. Example 2 - a convex mirror The same Star Wars action figure, 8.

Where is the image in this case, and what are the image characteristics? Again, the first step is to draw a ray diagram. This should tell you that the image is located behind the mirror; that it is an upright, virtual image; that it is a little smaller than the object; and that the image is between the mirror and the focal point. The second step is to confirm all those observations.

The mirror equation, rearranged as in the first example, gives: