What is an electric field and how is it created? To play it safe; we must provide TWO solutions: Solution I: using the mirror … (ii) distance between the insect and its image. 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 Draw a ray diagram to show the formation of the image in this case. 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. What will be the nature and the size of the image formed? How do you calculate the total resistance of a parallel circuit? Example 1: An object is placed in front of a plane mirror. The image formed at O’ is at a distance d behind the mirror. It is represented by the symbol m. Power of mirror Power of a mirror [in Diopters] = \(\frac { 1 }{ f(in\quad metre) } \). The image now formed at O” which is also at a distance d + x from M’. 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. Applications of Total Internal Reflection. How do you calculate the total resistance of a series circuit? Since, |v| = |u| = 2f, it means that the object is placed at the centre of curvature (c) of a concave mirror, the image formed is at the centre of curvature,real and inverted and of the same size as the object. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students. Now, magnification, ⇒ \(m=-\frac { v }{ u } =-\frac { 120 }{ 7\times (-40) } =+\frac { 3 }{ 7 } \) Since, the magnification is positive, the image is erect. Find the position of the image. Example 4: An object is placed at a distance of 15 cm from a concave mirror of focal length 10 cm. Calculate the following? Example 2: An insect is at a distance of 1.5m from a plane mirror. 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. 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. Definition : The ratio of the size of the image, as formed by reflection from the mirror to the size of the object, is called linear magnification produced by 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. 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 … Example 5: A 3 cm long object is placed perpendicular to the principal axis of a concave mirror. How is the Image Formed by a Spherical Mirror? 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. Solution: Suppose the object O was initially at a distance d from the plane mirror M as shown in fig. 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 . An object of size 7.0 cm is placed at a distance of 27 cm in front of concave mirror of focal length 18 cm. Giving reason answer the following: For the three object distances, identify the mirror\ mirrors … (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. Relationship between Energy Transferred, Current, Voltage and Time. Solution: We have u = –15 cm and f = –10 cm Which Type of Image is Formed by a Plane Mirror? An object of size 7 cm is placed at 27cm in front of a concave mirror of focal length 18 cm. An object 2 cm in size is placed 30 cm in front of a concave mirror of focal length 15 cm. Now, the mirror is shifted by a distance x to M’ such that the distance of the object from M’ becomes d + x. For a concave mirror, image size is same as the object size when it is placed at the center of curvature.

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