# what is magnification of mirror write its formula

## what is magnification of mirror write its formula

3 mins read. (xi) Thus the final position, nature and size of the image A'B' are:         (a) Position of image A'B' = 3.3 cm × 5 = 16.5 cm from the lens on opposite side. At what distance from the mirror should a screen be placed so that a sharp focussed image can be obtained? Delhi - 110058. Focal length, f = - 15 cm    [f is - ve for a concave lens]Image distance, v = - 10 cm [Concave lens forms virtual image on same side as the object, so v is - ve]As,                                                          Object distance, u = -30 cm. Quick summary with Stories. Hence it is a concave mirror. 5 min. Express m in terms of u, v and f. Linear magnification: The ratio of the height of the image to that of the object is called linear magnification or transverse magnification or just magnification. $$m = \frac{h_i}{h_o} = \frac{v}{u}$$ Converging lens means a convex lens. Problems Based on Mirror Formula and Magnification I. We are given a convex mirror. The mirror must be a spherical mirror (concave mirror) as the magnification in plane mirror is never -1 but always 1. The linear magnification or magnification of a spherical mirror may be defined as the ratio of the size (height) of the image to the size (height) of the object. Find the size and the nature of the image. $$m = \frac{h_i}{h_o} = \frac{v}{u}$$ Q. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. Calculating magnification with the help of lens formula: Magnification of a lens is defined as the ratio of the height of an image to the height of an object. Express m in terms of u, v and f. Linear magnification: The ratio of the height of the image to that of the object is called linear magnification or transverse magnification or just magnification. What will be the distance of the object, when a concave mirror produces an image of … The sign convention for spherical mirrors follows a set of rules known as the “New Cartesian Sign Convention”, as mentioned below: a. where,   v = image distance u = object distance. Another point to be noted is that if the value of magnification is equal to 1, then the image formed is of the same size as that of object. As the distances given in the question are large, so we choose a scale of 1: 5, i.e., 1 cm represents 5 cm. What do you meant by linear magnification? c. The object is always placed on the left side of the mirror which implies that light falling from the object on the mirror is on the left-hand side. Learn more about Reflection of Light here. (b) Nature of image A’B’: Real and inverted. Object distance is the distance of the object from the pole of the mirror; denoted by the letter u. Draw the ray diagram and find the position, size and the nature of the image formed. Ltd. Download books and chapters from book store. On the other hand, the real image is always inverted and formed below the principal axis so h2 will be negative. It is denoted by m. Here, we have Object size, h = + 5 cm Object distance, u = -20 cmRadius of curvature, R = + 3.0 cm [R is +ve for a convex mirror]∴ Focal length ,  f = R2 = +15 cm From mirror formula,                      1v = 1f-1u we have,                       1v= 1+15-1-20      = 4+360     = 760 Image distance, v = 607≃ 8.6 cm. The expression which gives t… © Magnification, m = -vu= h'h Therefore,  Image size, h' = -vhu                          = -8.6 × 5-20                         = 2.15 ≃ 2.2 cm. Besides, its formula is: Magnification (m) = h / h’ Here, h is the height of the object and h’ is the height of the object. It is equal to the ratio of image distance to that of object distance. How is linear magnification applicable in plane mirrors. The magnification of a mirror is represented by the letter m. Thus m = Or m = where, h 2 = size of image h 1 = size of object Download the PDF Question Papers Free for off line practice and view the Solutions online. Problems Based on Mirror Formula and Magnification- II. image formed is virtual and erect) and m will be negative when h2 is negative (i.e. unit. Calculating magnification with the help of lens formula: Magnification of a lens is defined as the ratio of the height of an image to the height of an object. It is also given in terms of image distance and object distance. Now, using the mirror formula,                      1u+1v = 1f∴                  1v = 1f-1u ⇒                      = 1-18-1-27 = -3+254 = -154i.e.,                  v = -54 cm The screen should be placed at a distance of 54 cm on the object side of the mirror to obtain a sharp image. b. m=\frac {v} {u} A virtual and erect image of height 2.2 cm is formed behind the mirror (because v is positive) at a distance of 8.6 cm from the mirror. It is denoted by m. But h2 can be positive or negative depending on whether the image formed is virtual or real. Problems on Spherical Mirrors. We are given a concave mirror. Ans. Problems Based on Mirror Formula and Magnification- I. The principal axis is taken as the x-axis of our coordinate system.