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Lethbridge College Applied Physics (PHY143) Unit 2 – Optics – Light and Mirrors The Nature of Light White light is made up of a range of electromagnetic waves with varying frequencies (or wavelengths). The electromagnetic spectrum may be divided into bands, as follows: Light has a constant speed (c) of 3.00 x 108 m/s. This speed is equal to the frequency of the electromagnetic wave (f with Hz as a unit, or seconds-1) times the wavelength (λ with meters as a unit). As one variable increases the other decreases, because c (the speed of light) is constant. Equation 2-1: c=fxλ You might recognize the labels that describe the electromagnetic bands, with radio and television waves to the far left of the spectrum (low frequency and large wavelength) to gamma rays (high frequency and small wavelength) to the far right of the spectrum. The visible spectrum is represented by a small band in the middle with a frequency range from 4.29 x 1014 Hz to 7.50 x 1014 Hz. The Law of Reflection When light hits a mirror or any reflective surface, it will be reflected. If the surface is perfectly flat, the light hitting the surface (the incident wave) will be reflected at the same angle relative to the normal (reflected wave). The reference plane that we call ‘normal’ is always perpendicular to the surface. Equation 2-2: Angle of Incidence = Angle of Reflection Θi = Θr If your surface is not perfectly flat, the light will diffuse. The light is still obeying the Law of Reflection, but the normal plane is always changing as the surface changes. It is important for specular reflection that the surface is not rough – the cost to produce a quality reflective surface is related in the costs of good optical equipment. Spherical Mirrors Take a hollow sphere and cut a circle from the surface. If this surface is reflective, you now have a spherical mirror. If the reflective surface is on the inside, the spherical mirror is concave; and if the reflective surface is on the outside, the spherical mirror is convex. When light reflects off of a concave spherical mirror, the rays will converge onto a single point – this is called the Focal Point. The focal point for a spherical mirror is always one half of the radius of the sphere. Equation 2-3 F=½C Images Using Spherical Mirrors – Graphical Method (Ray Tracing) If you were to look into a flat mirror, the image you see would be the same size as you, it would be upright, and it would be the same distance behind the surface of the mirror as you are standing in front of it. This is not true for spherical mirrors. To determine the location of the image, the size of the image, and the orientation of the image - whether the image is upright or inverted (up-side-down) - one can use either a graphical or analytical method. In other words, to determine the characteristics of the image in a mirror, you can either draw it or calculate it. It is extremely important, however, that you take the time to use the graphical method before using the analytical method - mainly because it is practically fool-proof. We call the graphical method to determine the characteristics of an image in a spherical mirror Ray Tracing. We use three rays (from the infinite number actually coming from the visible object) to help solve mirror design problems: these are the Parallel ray, the Focal-point ray, and the Centre ray. For convex or concave mirrors, we must know the focal point (or the radius of the sphere). To scale, we must draw the object that is being reflected by the mirror, using the correct height of the object at the correct distance from the surface of the mirror. We may then draw the three important rays – where they intersect indicates the location, size, and orientation of the image. Convex Mirrors For example, note the following convex mirror: The object stands in front of a convex mirror. Consider first the P-ray: it leaves the top of the object as a parallel ray, and when it hits the convex mirror it is reflected. The reflected ray passes through the focal point of the spherical mirror. Even though the ray would be reflected up and away to the left in reality, there is a ‘virtual’ direction for the ray which passes through the focal point. Now consider the C-ray: it leaves the top of the object and travels straight through the mirror towards the center of the sphere. In reality, this ray would reflect directly back on the path that it arrived, as it is hitting the surface along the normal plane. The ‘virtual’ path of the ray will pass directly through the center of curvature for the mirror. And finally, consider the F-ray: it leaves the top of the object and travels towards the focal point of the mirror. When it hits the surface of the mirror it is reflect parallel to the datum. In reality the ray is reflected back towards the object, but a ‘virtual’ ray passes horizontally through the mirror. The point where these three rays intersect is the top of the image. In this case, the image is closer to the surface of the mirror than the object, it is smaller than the object, and the image is upright. Because the image appears to be inside or behind the surface of the mirror, it is called a Virtual Image. The image from a convex mirror will always be upright, reduced in size, and virtual. Summary Object Size Orientation Type Anywhere Smaller Upright Virtual Concave Mirrors Now, let’s trace the rays using a concave mirror. As with the convex mirror, the parallel ray travels horizontally to the surface of the mirror, and is then reflected through the focal point of the mirror. The F-ray travels through the focal point of the mirror to the surface where it is reflected parallel to the datum. And the C-ray passes directly through the center of curvature for the mirror (the radius of the sphere). The image will be located at the point where the three rays intersect. In this case the image is smaller than the object, closer to the surface of the mirror than the object, and inverted. The image is called a Real Image because it is located on the same side of the mirror as the object. What would happen to the image if we moved it closer to the mirror – for example, between the center and the focal point? When the object is between the center of radius and the focal point, the image will be larger than the object, inverted, and will appear farther away from the surface of the mirror than the object. The image in this example is ‘real’. If we move the object even closer to the mirror … Self Check How would you describe this image? For a Concave Mirror, you can expect the following image characteristics depending where the object is located: Summary Object Size Orientation Type Beyond C At C Between C & F Just beyond F Just inside F Between F & surface Smaller Same Larger Nearing infinite Nearing infinite Larger Inverted Inverted Inverted Inverted Upright Upright Real Real Real Real Virtual Virtual Images Using Spherical Mirrors – Analytical Method There are two formulae that are used to solve optical problems analytically. These are: Equation 2-4: The Mirror Equation 1 + 1 = 1 di do F where: di do F = distance from mirror to the image = distance from the mirror to the object = focal point Equation 2-5: Magnification Equation m where: hi ho m = hi ho = -di do = height of the image = height of the object = magnification The reason we encourage you to use the graphical method to estimate the size, location and orientation of the image is that it is quite easy to make errors using the analytical method. The most common source of error is using the correct signs for the variables. The following table outlines the sign rules. Focal length F is positive for concave mirrors F is negative for convex mirrors Magnification m is positive for upright images m is negative for inverted images Image Distance di is positive for real images di is negative for virtual images (Remember, real images are images that are located on the same side of the mirror as the object, and virtual images appear to be located behind the surface of the mirror) Object Distance do is positive for objects in front of a mirror (real objects) do is negative for objects behind the surface of the mirror (virtual objects) (Note: do is typically found in multiple mirror systems in which one mirror is reflecting a virtual image to another mirror) EXAMPLE – Concave Mirror Using a concave mirror with a focal point of 25 cm, we would like to create a virtual image that is twice the size of the original object. The original object is 12 cm high. First Step – think about the problem … Looking at the summary chart for a concave mirror, to create an enlarged virtual image the object will be located between the focal point and the mirror. We know we can use a concave mirror for this application. Second Step – extract the variables from the problem (there are only six possible variables). F di do hi ho m = 0.25 m =? =? =? = 0.12 m =2 Third Step - check signs F is positive for concave mirrors. m is positive for upright images (we know it is upright from the summary table). di is negative for virtual images (we know it is virtual from the summary table). do is positive because it is a real object. Fourth Step - solve analytically m = -di/do +2 = - di /do … di = - 2do Substitute into the Mirror Equation 1/do + 1/di = 1/F 1/do + 1/(-2do) = 1/(0.25) … 1/do – 0.5/do = 4 0.5/do = 4 do = 0.125 m (the object is located half way between the focal point and the surface of the mirror. This seem reasonable if you consider our thoughts in Step 1) Fifth Step - use the graphical method to check your answer. By locating the object at 0.125 m (indicated by the analytical solution), the image will be twice the size, upright and virtual. We can be confident that our solution is correct. Self Check – try the following problems on your own. 1. You notice a convex mirror mounted on the ceiling at your local organic food store. The clerk tells you that she cut the mirror from a sphere with a diameter of 2 m. The store is roughly 10m wide and 25 m long, and the ceiling is approximately 4 m high. Your image in the mirror is about 10cm high and you are 1.85 m tall. How far away are you from the mirror. (Caution: you don’t need all of the information I have provided to solve the problem). 2. An box that is 35 cm high, 20 cm wide and 15 cm long sits 28 cm from a concave mirror. A real image is formed with a height of 47 cm. What is the focal point of the mirror? 3. You are driving and look in the side mirror which states: “Objects appear closer than they are”. The virtual image is upright and is approximately 5% of the size of the actual object. It appears to be 15 m behind the surface of the mirror. How far behind you is the real object? Self Check Solutions 1. In the convex mirror your image is smaller, upright and virtual. Referring to the summary table for a convex mirror, this seems reasonable. F do di hi ho m =½C =? =? = 0.10 m = 1.85 m =? = ½ (2 m / 2) = 0.5 m Checking signs: F is negative for convex mirrors. m is positive for upright images di is negative for virtual images do is positive for real objects Using the Magnification Equation: hi/ho = -di/do … do = -di * ho / hi do = - di (1.85) / (0.1) … do = -18.5 di Using the Mirror Equation: 1/do + 1/di = 1/F 1/ (-18.5 di) + 1/ di = 1/(-0.5) … 0.946 / di = -2 di = -0.47 m (a negative answer is what we would expect from the sign rules) do - -18.5 * (-0.47) = 8.75 m You are standing roughly 8.8 m from the convex mirror. Check by drawing this scenario. The graphical solution shows a virtual, upright image at approximately 10 cm compared to the original object with a height of 1.85 m. The virtual image is located near the focal point at about 0.5 m behind the surface of the mirror. 2. A concave mirror that produces a larger, real image will require that the object be located between the C and F for the mirror. The image will be inverted according to the summary table for a concave mirror. F do di ho hi m =? = 0.28 m =? = 0.35 m = 0.47 m =? Checking signs: F is positive for concave mirrors. m is negative for inverted images (therefore hi will have a negative sign) di is positive for real images do is positive for real objects Using the Magnification Equation: hi/ho = -di/do … di = -do * hi / ho di = - (0.28) * (-0.47) / (0.35) … di = +0.376 m Using the Mirror Equation: 1/do + 1/di = 1/F 1/ (0.28) + 1/ (0.376) = 1/F … 6.231 = 1/F F = 0.16 m (a positive answer is what we would expect from the sign rules) The mirror you require will have a focal point of 16 cm. Check by drawing this scenario. The image shown is real, inverted, and larger than the original object, as expected. 3. In the convex mirror your image is smaller, upright and virtual. Referring to the summary table for a convex mirror, this seems reasonable. F do di hi ho m =? =? = 15 m =? =? = 5% = 0.05 Checking signs: F is negative for convex mirrors. m is positive for upright images di is negative for virtual images do is positive for real objects Using the Magnification Equation: m = -di/do … do = -di / m do = - (-15) / (0.05) … do = 300 The real object is 300 m in front of the surface of the mirror. Check by drawing this scenario. The graphical solution shows a virtual, upright image at considerably smaller than the original object. If you were to draw this to scale the results would closely match the solution found using the analytical method.