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W.S #4 Image formation - Mirrors - Ray diagrams & Eqn Name:_______________ A curved mirror is typically a cutout of a spherical mirror. If the cutout is such that the reflective coating is on the outside surface so light reflects from the inside of this imaginary sphere, we call that mirror a concave mirror. If the reflective coating is on the inside so light reflects off the outside surface, we call that mirror a convex mirror. ● ● ● A line Loading… to the mirror surface through the center of the mirror is called the principal axis. A curved mirror has symmetry about this axis. The center of curvature is the point on the principal axis which coincides with the center of the sphere from which the mirror is cut out of The focus or focal point is the point where all rays parallel and close to the principal axis are reflected through. Focus Loading… is a distance half the distance of the center of curvature Loading… from the point where the principal axis meets the mirror. Rules of ray tracing through curved mirrors - object with one end on the principal axis - typical rays are drawn from the other end of the object: 1. An incident ray through the center of curvature reflects right back through the center of curvature. 2. An incident ray parallel to the principal axis reflects through the focus or focal point. 3. An incident ray pointing where the principal axis meets the mirror reflects symmetrically on the other side of the principal axis. You only need two rays to find the image. Choose any two of the above. Stick to those for all diagrams. Note that sometimes, you might have to extend a ray backward in order to satisfy the above requirements. Let’s practice ray tracing for finding images for a concave mirror: (A) (B) (C) Ray tracing for a convex mirror: (D) (E) (F) Classification of images: Upright or inverted? As we have seen, plane mirrors and convex mirrors produce an upright image. This is an image that is in the same orientation as the object. An inverted image, like that shown, is one in which the image is upside down in relation to the object. Real or virtual? ● ● ● ● ● ● A virtual image is one that the light does not actually pass through. Instead, our brains see an image there because, when we look in the mirror at the object, our brains are so used to light traveling in straight lines that we trace all the reflected rays back to their apparent source, the point behind the mirror where the light appears to come from. A real image is formed when the reflected rays actually converge and meet to form the image. Real images can be captured on a screen like a camera; virtual images can not! Real images appear on the same side of the mirror as the object; Virtual images appear on the opposite side ‘behind’ the mirror For a single mirror, when the image is virtual it is also upright. For a single mirror, when the image is real it is also inverted. Larger or smaller? A plane mirror, as we investigated earlier, produces an image that is always the same size as the object. Convex (diverging) mirrors, on the other hand, always produce an image that is smaller than the object. Concave (converging) mirrors, can produce an image that is larger, smaller, or the same size as the object. Quantifying Images formed by mirrors: MAGNIFICATION Consider a ray incident in the principal axis as shown. The two right triangles formed above and below the principal axis are similar. Hence | ℎ𝑖 | | 𝑠𝑖 | | ℎ | =| 𝑠 | | 𝑜| | 𝑜| where ℎ is the height and 𝑠 is the distance from the |ℎ | |𝑠 | mirror, 𝑖 for image and 𝑜 for object. The quantity | ℎ 𝑖 | = | 𝑠 𝑖 | = 𝑚 is called | 𝑜| | 𝑜| the magnification of the image. It is NOT a fixed quantity for a mirror, rather it changes depending on the type of mirror and the object location. NOTE ABOUT SIGN CONVENTION: In this case shown, ℎ𝑖 is a negative quantity (below the axis - inverted image). An upright object height ℎ𝑜is positive. The object distance 𝑠𝑜 is ALWAYS Positive. Image distances 𝑠𝑖 is positive for a real image and negative for a virtual image. DO THIS: Go back to ray diagrams (A) and (D) and calculate the magnification using a ruler using both equations. Quantifying Images formed by mirrors: IMAGE LOCATION Consider a ray parallel to the principal axis that reflects through the focus. The triangles above and below the principal axis are similar. Noting that the height ℎ𝑖 is a negative quantity (below the axis - inverted image), we have ℎ 𝑠 −𝑓 − 𝑖 𝑖 = ℎ0 Simplying 𝑓 𝑠𝑖−𝑓 𝑓 = = 𝑠𝑖 𝑠0 𝑠𝑖 𝑠0 ⇒ 𝑠𝑖 𝑓 − 1 = 𝑠𝑖 𝑠0 Dividing both sides by 𝑠𝑖 gives 1 𝑓 − 1 𝑠𝑖 = 1 𝑠𝑜 → 𝑀𝑖𝑟𝑟𝑜𝑟 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 𝑓 = 1 𝑠𝑜 + 1 𝑠𝑖 SIGN CONVENTION (repeated): ● Focal length 𝑓 is positive for a concave mirror (because it lies on the ‘real’ side); negative for a convex mirror (because it lies on the virtual side) ● Height is positive for an upright image and negative for an inverted image ● Object distance 𝑠𝑜 is always positive; Image distance 𝑠𝑖 is positive for a real image and negative for a virtual image. PRACTICE PROBLEMS: ALWAYS draw a ray diagram to check if your answer is reasonable. #1: A concave mirror has a focal length of 18 cm. (i) Where will an image form if an object is placed 58 cm from the mirror? (ii) If the object is 12 cm tall, what will be the height of the image? Is the image erect or inverted? _________ real or virtual? __________ #2: (i) Where will the image form if the same object is placed 36 cm in front of the same mirror as in #1? (ii) How tall is the image? Is the image erect or inverted? _________ real or virtual? __________ #3: (i) Where will the image form if the same object is placed 32 cm in front of the same mirror as in #1? (ii) How tall is the image? Is the image erect or inverted? _________ real or virtual? __________ #4: (i) Where will the image form if the same object is placed 6.0 cm in front of the same mirror as in #1? (ii) How tall is the image? Is the image erect or inverted? _________ real or virtual? __________ #5: (i) Where will the image form if the same object is placed 15 cm in front of a convex mirror with a focal length of 18 cm? (Hint: The focal length is behind the mirror.) (ii) How tall is the image? Is the image erect or inverted? _________ real or virtual? __________ # 6: #7: #8: In a particular case of an object in front of a spherical mirror, the object distance is 12 cm and the magnification is + 4.0. Find (a) the image distance, (b) the mirror’s focal length. #9: A model of a horse is placed 32 cm away from a mirror that has a focal length of –20 cm. The model is 5.0 cm tall. Determine (a) the location of the image, (b) the height of the image, (c) whether the image is real or virtual, and (d) whether the image is upright or inverted. #10: In the situation shown in the Figure on the right, a small red LED (light-emitting diode) is placed on the principal axis 4.0 cm from a concave mirror that has a radius of curvature of 14 cm. The LED can be considered to be a point source which emits rays in all directions just like an object. Find the location of the image of the LED using equations and check against a ray diagram.