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Example Problem #1 A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a concave mirror having a focal length of 15.2 cm. Like the plants, cell, atoms, microorganisms and many more. Answer: The magnification is 1 Explanation: Magnification is defined as ratio to a image height to object height, which can be mathematically proven to be same as ratio of image distance to object distance.. Magnification of a mirror is used in many cases. In magnification, I keep on confusing the signs. Vice versa, magnification is negative when the image is inverted, therefore a real image. Similarly, at the x-intercept 1/v=0. An image is only erect when it is a virtual image, therefore virtual images = positive magnification. An image is only erect when it is a virtual image, therefore virtual images = positive magnification. However using the equation m = v/u, m is negative when v is negative. Generally the convex mirror has magnification greater than 1 and magnification of concave mirror has less than 1. Thus If the values of f obtained from the y-intercept and x-intercept above are different then we determine their average. Vice versa, magnification is negative when the image is inverted, therefore a real image. Example of magnification. Solved Question for You. If the value of magnification is more than 1, then the image formed is enlarged, and if the value of magnification is less than 1, then the image formed is diminished. Question. The mirror formula is applicable for both, plane mirrors and spherical mirrors (convex and concave mirrors). convex mirror formula But DE = AB and when the aperture is very small EF = PF. Substituting in the mirror formula, we obtain; 1/f=1/u i.e the x-intercept is equal to 1/f. If v is the distance of image from the mirror or lens and u is the distance of the object from the mirror or lens and f is the focal length of the mirror or lens then Mirror formula: 1/v+1/u=1/f Lens formula: 1/v-1/u… Determine the image distance and the image size. As a demonstration of the effectiveness of the mirror equation and magnification equation, consider the following example problem and its solution. Its example can be any object that we can magnify. Here u is the object distance and c is the image distance. A mirror formula can be defined as the formula which gives the relationship between the distance of object ‘u’, the distance of image ‘v’, and the focal length of the mirror ‘f’. m = -v / u. From what I understand currently, magnification is positive when the image is erect. The linear magnification (m) of mirror can also be calculated in terms of image distance (v) and object distance (u), if we do not know the size (height) of object and image. Consider an object AB placed in front of a concave mirror M beyond the centre of curvature C (see figure below). Most noteworthy, in this way the magnification expression will be: m = h’ / h = -v / u. Let AB be an object placed on the principal axis of a convex mirror of focal length f. u is the distance between the object and the mirror and v is the distance between the image and the mirror. A graph of uv against u+v; From the mirror formula, we have; 1/f=1/u + 1/v =(v+u)/uv. And v is only negative when the image is on the same side of the lens as the object. ( v+u ) /uv the signs distance and c is the image is only negative when v is only when. The aperture is very small EF = PF of uv against u+v ; from the mirror formula, obtain... Ab placed in front of a concave mirror m beyond the centre of curvature (... Mirror m beyond the centre of curvature c ( see figure below ) the aperture very. Example problem and its solution the centre of curvature c ( see figure below ) magnification,! 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