(iv) Join any point D (nearly at the top of lens) and F by a dotted line. Magnification is also equal to the ratio of image distance to the object distance. (iii) Mark points F and B on the left side of lens at a distance of 3 cm and 2 cm respectively. The magnification, m produced by a spherical mirror can be expressed as: m=\frac {h} { {h}’} Here, h is the height of image and h’ is the height of the object. It is equal to the ratio of image distance to that of object distance. m=\frac {v} {u} A concave lens of focal length 15 cm forms an image 10 cm from the lens. Download the PDF Question Papers Free for off line practice and view the Solutions online. Find the position of the image, its nature and size. Problems Based on Mirror Formula and Magnification I. What is its S.I. It is also given in terms of image distance and object distance. image formed is real and inverted). The principal axis is taken as the x-axis of our coordinate system. (b) For a convex mirror: Using cartesian sign convention, we find from the given fig that A' B' = + h' (Upward image height)AB = + h (Upward object height)PB' = + v (Image distance on right)BP = - u (Object distance on left)From similar triangles A' B' P and ABP, we get Linear magnification in terms of u and fi From mirror formula, we have or Linear magnification in terms of v and f: Again, from mirror formula, we have Hence, for any spherical mirror, concave or convex, we have. 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. b. Hence, the expression for magnification (m) becomes: m = h’/h = -v/u. How is linear magnification applicable in plane mirrors. should be placed at a distance of 54 cm on the object side of the mirror to obtain a sharp image. Ltd. Download books and chapters from book store. Define magnification for a spherical mirror. The Mirror formula explains how object distance (u) and image distance (v) are related to the focal length of a spherical mirror. Furthermore, the letter ‘m’ denotes the magnification of the object. 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. Ans. 6 min. Besides, its formula is: Magnification (m) = h / h’ Here, h is the height of the object and h’ is the height of the object. (vi) Draw a line A'B', perpendicular to principal axis from B'. We draw the ray diagram as follows:(i) Draw the principal axis (a horizontal line). We are given a convex mirror. The sign convention for spherical mirrors follows a set of rules known as the “New Cartesian Sign Convention”, as mentioned below: a. At what distance from the mirror should a screen be placed so that a sharp focussed image can be obtained? 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. Test Your Understanding and Answer These Questions: Rules for Obtaining Images by Spherical Mirrors. 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. The mirror must be a spherical mirror (concave mirror) as the magnification in plane mirror is never -1 but always 1. 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]. (v) Draw a line AD, parallel to principal axis. Find the size and the nature of the image. https://www.zigya.com/share/U0NFTjEwMDUyNTI0. 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 But h2 can be positive or negative depending on whether the image formed is virtual or real. (iii) Mark two foci F and F' on two sides of the lens, each at a distance of 2 cm from the lens. Therefore, on this scale 5 cm high object, object distance of 25 cm and focal length of 10 cm can be represented by 1 cm high, 5 cm and 2 cm lines respectively. Problems Based on Mirror Formula and Magnification- II. What do you meant by linear magnification? Besides, its formula is: Magnification (m) = h / h’ Here, h is the height of the object and h’ is the height of the object. Image distance is the distance of the image from the pole of the mirror and it is denoted by the letter v. And focal length is the distance of the principal focus from the pole of the mirror. The formula of magnification represents the ratio of the height of the image to the ratio of the height of the object. Also convex mirror always produces images smaller than the size of the object so it's magnification is always less than 1. 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. (vii) Let the two lines starting from A meet at A'. It will be found to be equal to 6 cm.Thus, object is placed at a distance of 6 cm × 5 = 30 cm from the lens. 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. Problems on spherical mirrors. The formula of magnification represents the ratio of the height of the image to the ratio of the height of the object. Magnification is the increase in the image size produced by spherical mirrors with respect to the object size. The pole (p) of the mirror is taken as the origin. The image is real, inverted and enlarged in size. Thus magnification m will be positive when h2 is positive (i.e. 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 Problems on Spherical Mirrors. Write the expression for magnification for (i) concave mirror (ii) convex mirror. Solved Example for You. Learn more about Reflection of Light here. It is equal to the ratio of image distance to that of object distance. We are given a concave mirror. Here, Object size, h = + 7.0 cmObject distance, u = - 27 cmFocal length, f = - 18 cm Image distance, v = ? As the virtual image is always erect and above the principal axis therefore h2 will be positive. (ix) The AB is position of object. VIEW MORE. (b) Nature of image A’B’: Real and inverted. (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. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. It is found that CB' = 3.3 cm and A'B' = 0.7 cm. It is the ratio of the height of the image to the height of the object and is denoted as m. The magnification, m produced by a spherical mirror can be expressed as: (vii) Draw a line CA', backwards, so that it meets the line from D parallel to principal axis at A. It is denoted by m. Problems Based on Mirror Formula and Magnification- I. Magnification can also be related to the image distance and object distance; therefore it can also be written as: m = -v/u. (c) Height of image A'B': 0.7 × 5 = 3.5 cm, i.e., image is smaller than the object. Find Magnification Given Object Distance = u = –3 Image Distance = v = –12 So Magnification m is given by m = (−)/ m = (−(−12))/((−3)) m = (−12)/3 m = −4 u and v are negative because they are in front of the mirror 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. \( m = \frac{h_i}{h_o} = \frac{v}{u} \) (ii) Draw a convex lens, keeping principal centre (C) on the principal axis. Where v is the image distance and u is the object distance. e. All the distances … 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. 3 mins read. There is no unit of magnification (m) as it is the ratio of two similar quantities. Draw the ray diagram. 2020 Zigya Technology Labs Pvt. Delhi - 110058. It is also given in terms of image distance and object distance. As the distances given in the question are large, so we choose a scale of 1: 5, i.e., 1 cm represents 5 cm. It is denoted by m. For a concave mirror: Using carteisn sign convention, we find from the given fig, that, From similar triangles A' B' P and ABP, we get, Linear magnification in terms of u and fi From mirror formula, we have, Linear magnification in terms of v and f: Again, from mirror formula, we have, Hence, for any spherical mirror, concave or convex, we have.