the focal length of a concave mirror is 30 cm
Mirror formula Definition : The equation relating the object distance (u) the image distance (v) and the mirror focal length (f) is called the mirror formula. Solution: The radius of curvature of the mirror = 30 cm Thus, the focal length of the mirror =\(\frac { 30 cm }{ 2 } \) = 15 cm. Solution: We have u = –15 cm and f = –10 cm Using the relation, \(\frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ f }\) we get \(\frac { 1 }{ v } +\frac { 1 }{ -15 } =\frac { 1 }{ -10 } \) or \(\frac { 1 }{ v } =\frac { 1 }{ 15 } -\frac { 1 }{ 10 } =-\frac { 1 }{ 30 } \) or v = –30 cm So the image will be formed 30 cm from the mirror. The minus sign shows that it is on the lower side of the principal axis, i.e. How is the Image Formed by a Spherical Mirror? (iii) Since h’ is positive, the image will be on the same side of the principal axis as the object. So, screen should be placed at a distance of 30 cm on the same side of the object in order to obtain a sharp image. At t = 0 the shape of the mirror begins to… Draw a ray diagram to show the formation of the image in … Solution for An object of height 79.2 cm is distance so = 80.4 cm from a concave mirror of focal length 9.62 cm. Filed Under: Physics Tagged With: Focal Length related to Radius of Curvature, linear magnification For spherical mirrors, Mirror formula, Power of mirror, ICSE Previous Year Question Papers Class 10. Solution: (i) The distance of insect from the mirror = 1.5 m ∴ The distance of insect from the mirror is also equal to 1.5 m. The image is formed at 1.5 m behind the mirror. So, OM = MO’ = d OM’ = M’O” = d + x Thus, OO” = OM’ + M’O” = 2(d + x) …(1) when OO’ = OM + MO’ = 2d …(2) ∴ O’O” = OO” – OO’ = 2(d + x) – 2d = 2x Thus, the image is shifted from O’ to O” by a distance 2x. Calculate the focal length of the mirror. Calculate the following : (i) location of the image (ii) height of the image (iii) nature of the image Solution: (i) For a convex mirror, focal length is positive. The image formed at O’ is at a distance d behind the mirror. Example 2: An insect is at a distance of 1.5m from a plane mirror. An object is placed at a distance of 30 cm in front of a convex mirror of focal length 15 cm .find the nature and position of image ? in front of the mirror as shown in fig. What do you mean by Total Internal Reflection? How do you calculate the total resistance of a parallel circuit? (ii) Magnification, or \(m=\frac { h’ }{ h } =-\frac { v }{ u } \) \(\frac { h’ }{ h } =\frac { (-6) }{ (-10) } =+0.6 \) or h’ = + 0.6 × h0 = 0.6 × 1.4 = 0.84 cm. Example 4: An object is placed at a distance of 15 cm from a concave mirror of focal length 10 cm. Giving reason answer the following: For the three object distances, identify the mirror\ mirrors … To play it safe; we must provide TWO solutions: Solution I: using the mirror … (ii) The distance between the insect and image = 1.5 + 1.5 = 3m, Example 3: A concave mirror is made up by cutting a portion of a hollow glass sphere of radius 30 cm. Which Type of Image is Formed by a Plane Mirror? Applications of Total Internal Reflection. Determine the focal length of a concave mirror if: (a) with the distance between an object and its image being equal to l = 15 cm, asked Dec 9, 2018 in Physics by pinky ( 74.2k points) optics Calculate the height of the image formed. Is an Image formed by Reflection Real or Virtual, Reflection of Light from Spherical Mirror. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. An object 2 cm in size is placed 30 cm in front of a concave mirror of focal length 15 cm. What is the Relationship between Electric Current and Potential Difference? An object 2 cm in size is placed 30 cm in front of a concave mirror of focal length 15 cm. Thus, the height of the image is 0.84 cm. Solution: We have u = –15 cm and f = –10 cm The incident rays make small angles with the mirror surface or the principal axis. What is alternating current and direct current? v=? f=20 cm. Draw a ray diagram to show the formation of the image in this case. the image is inverted. What will be the nature and the size of the image formed? It does not say the NATURE of the image. The image now formed at O” which is also at a distance d + x from M’. An object of size 7 cm is placed at 27cm in front of a concave mirror of focal length 18 cm. Solution: The radius of curvature of the mirror = 30 cm Thus, the focal length of the mirror =\(\frac { 30 cm }{ 2 } \) = 15 cm. Find the position of the image. Since radius of the mirror is twice the focal length, center of curvature lies at a distance of 30 cm from the pole. What is an electric field and how is it created? Since n has a negative sign, the image is formed to the left of the mirror, i.e. Solution: We have u = –15 cm and f = –10 cm Thus, the image is formed 17.1 cm behind the mirror. Relationship between Energy Transferred, Current, Voltage and Time. Find the position of the image. Now, the mirror is shifted by a distance x to M’ such that the distance of the object from M’ becomes d + x. behind the mirror. For a concave mirror, image size is same as the object size when it is placed at the center of curvature. Solution: Here u = –15 cm and n = –30 cm Size of the object, h = 2 cm Magnification, \(m=\frac { h’ }{ h } =-\frac { v }{ u } \) or \(\frac { h’ }{ h } =-\frac { (-30) }{ (-15) } =2 \) or h’ = – 2 × h = – 2 × 3 = – 6 cm So the height of the image is 6 cm. Damped Oscillations, Forced Oscillations and Resonance. Example 1: An object is placed in front of a plane mirror. Hence, an object kept at 30 cm will produce an image of the same size as that of the object. How do you calculate the total resistance of a series circuit? What will be the nature and the size of the image formed? Calculate the following? The distance of the object from the mirror is 15 cm, and its image is formed 30 cm from the mirror on the same side of the mirror as the object . For each concave mirror you perform the experiment of image formation for three values of object distance of 10 cm, 20 cm and 30 cm. Therefore, f = +15 cm and u = –10 cm Using the relation, \(\frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ f } \), we get \(\frac { 1 }{ v } +\frac { 1 }{ -10 } =\frac { 1 }{ 15 } \) or \(\frac { 1 }{ v } =\frac { 1 }{ 15 } +\frac { 1 }{ 10 } =\frac { 5 }{ 30 } =\frac { 1 }{ 6 } \) or v = 6 cm Since v is positive, the image is formed to the right of the mirror at a distance 6 cm from it. If the mirror is moved away from the object through a distance x, by how much distance will the image move? An object 2 cm in size is placed 30cm in front of a concave mirror of focal length 15cm. The object lies close to principal axis of the mirror. A converging beam of solar rays is incident on a concave spherical mirror whose radius of curvature is 0.8 m. Determine the position of the point on the optical axis of the mirror where the reflected rays intersect, if the extensions of the incident rays intersect the optical axis 40 cm from the mirror's …
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