as for mirror formula . Another way to prevent getting this page in the future is to use Privacy Pass. From 1/u – 1/v graph : We can also measure the focal length by plotting graph between 1/-u and 1/v. Plug in f = 0.1m . The graphs between 1/v and 1/u for concave and convex mirror are shown below. v versus u graph for convex and concave mirror. Simulator Procedure (as performed through the Online Labs) Select the focal length using the slider. (ii) from u - v graph (iii) from 1/u - 1/(v ) graph. The focal length of the mirror can also be measured graphically by plotting graphs between u & v, and 1/u & 1/v. ⇒ y = c + x , this is represented by graph (iii). Thus option (c) is correct. Performance & security by Cloudflare, Please complete the security check to access. i/f = u-v/uv u =distance of object needle from optical center of the lens (-) f =uv/u-v v = distance of image needle from optical from optical center of lens (+) Procedure . Thus option (c) is correct. Theory : The relation between u , v and for a convex lens is. 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. 1/v and 1/u Graph . The 1 ∕u versus 1 ∕v graph is shown in the figure. Download PDF for free. Since convex mirror always produced virtual image, option (d) is not correct. (a) plane mirror (b) concave lens (c) convex lens (d) concave mirror. The parallel rays from the outer edge are deviated towards the middle in a: (a) convex mirror (b) concave lens (c) concave mirror (d) convex lens, The eye defect ‘presbyopia’ can be corrected by: (a) convex lens (b) concave lens (c) convex mirror (d) Bi focal lenses, For a real object, which of the following can produce a real image? Let f = 10 cm = 0.1m. Your IP: 172.105.117.219 We can repeat the experiment with concave mirrors of different focal lengths. The graphs between object distance ‘u ’ and magnification ‘ m ’ for concave and convex mirror are given as. Two types of spherical mirrors are; Concave mirror: Its inner concave surface reflects, and has polished outer surface. Use mirror formula we get. The relation between u 1 and v 1 is a linear relation, and . If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. Use mirror formula 1/v + 1/u =1/f. For virtual image v > 0 ,u < 0 and f > 0 . Plot of object distance vs image distance for spherical mirrors - diagram. • • v+ u = … 1/u + 1/v = 1/f ⇒ 1/v = 1/f - 1/u… (a) Graph (i) represent for virtual image formed by convex mirror, (b) Graph (ii) represent for virtual image formed by concave mirror, (c) Graph (iii) represent for virtual image formed by convex mirror, (d) Graph (i) represent for real image formed by convex mirror. The graph is a straight line intercepting the axes at A and B. Simulator Procedure (as performed through the Online Labs) Calculate the mean of all focal lengths to get the correct focal length of the given concave mirror. Hello, For convex mirror . Statement II: m = -v/u and convex mirror always forms virtual image. Let 1/v = y, 1/f = c and 1/u = -x ⇒ y = c + x , this is represented by graph (iii). For virtual image v > 0 ,u < 0 and f > 0 . Share with your friends. An object is placed at a distance u from a concave mirror and its real image is received on a screen placed at a distance of v from the mirror. If f is the focal length of the mirror, then the graph between 1/v versus 1/u is we get v +u = 10uv . Which of the following statements are correct? An object is placed at a distance u from a concave mirror and its real image is received on a screen placed at a distance of v from the mirror. Note that 1 ∕u is zero when the object is placed at infinity. The focal length of the mirror can also be measured graphically by plotting graphs between u & v, and 1/u & 1/v. It is a mirror which has the shape of a piece cut out of a spherical surface. If f is the focal length of the mirror, then the graph between 1/v versus 1/u is The graphs between 1/v and 1/u for concave and convex mirror are shown below. Plot a graph with 1/u along X axis and 1/v along Y axis by taking same scale for drawing the X and Y axes. ⇒ y = -c + x, this is represented by graph (ii). Theory Spherical mirrors. i want graphs b/w v VS u and 1/v VS 1/u for both concave and convex mirror also i want graphs b/w m VS v m VS u for both concave and convex mirror with explanation - Physics - … Statement I : Magnification of a convex mirror is always positive . u 1 + v 1 = f 1 and f is constant. Look at the ray diagram shown in the figure. An object is placed at a distance u from a concave mirror and its real image is received on a screen placed at a distance of v from the mirror. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. Thus option (b) is correct. 1/f= 1/v – 1/u Where f=focal length of convex lens. If f is the focal length of the mirror, then the graph between 1/v versus 1/u is For Concave mirror. Since convex mirror always produced virtual image, option (d) is not correct. In this case, the image is formed at the focus i.e., v = f. SVG-Viewer needed. For concave mirror. Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Cloudflare Ray ID: 5f998e8eca8dcc10 For convex mirror . Let f = -10 cm = -0.1m. If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. Share 4. You may need to download version 2.0 now from the Chrome Web Store. For virtual image v > 0, u < 0 and f > 0. The graph is a straight line with slope -1 and intercept on the 1 ∕v-axis is 1 ∕f. We can repeat the experiment with concave mirrors of different focal lengths. 1.2 Convex Lens. Mount the concave mirror in the mirror holder . For a convex lens v 1 − u 1 = f 1 v u u − v = f 1 (u − v) (v u) = f The above equation looks like a hyperbola.