1st problem: A small object of size 0.5 mm is placed close to a convex lens of focal length 5cm. The virtual image formed is at a distance of 25 cm from the lens. Find magnification. Data: focal length, f = 5cm distance of image , q = -25 cm (because image is virtual) distance of object , p = ? magnification, M = ? 1/f = 1/p + 1/q M = height of image/height of object = q/p (ratio between height and distance of image and object) Solution: First finding p, 1/f = 1/p + 1/q 1/p = 1/f - 1/q 1/p = 1/5 - (1/-25) .................. 1/p = 6/25 p = 25/6 = 4.16cm Now putting value of 'p' in magnification formula, M = height of image/height of object = q/p (ratio between height and distance of image and object) M = -25/4.16 M = -6 Can magnification be -ve? It is so in this problem. What have I done wrong? 2nd problem: 1-This part of a numerical problem as our teacher did it in classroom. Data: distance of object, p = 20cm focal length of concave lens,f = -19 (Is focal length of concave lens really always -ve?) distance of image, q = ? Solution: According to lens formula, 1/f = 1/p + 1/q => 1/q = 1/f - 1/p 1/q = 1/-19 - 1/20 1/q = -20-19/380 1/q = -39/380 q = -9.74cm Why can't LCM be negative at line 4,5 of solution when one of the denominators is -ve? And is the -ve value of distance of image correct?