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Q. 26. An object of height 4.0 cm is placed at a distance of30 cm from the optical centre 'O' of a convex lens offocal length 20 cm. Draw a ray diagram to find theposition and size of the image formed. Mark opticalcentre 'O' and principal focus 'F' on the diagram.Also find the approximate ratio of size of the imageto the size of the object. (2)5.04.54.03.5

Answer»

Height of object, h₀ = 4cm object distance from lens , u = 30cm Focal length of lens , f = 20cm

Now, use formula, 1/v - 1/u = 1/f Here, u = -30cm, f = 20cm ∴ 1/v -1/-30 = 1/20 ⇒ 1/v = 1/-30 + 1/20 = (30 - 20)/20×30 = 1/60⇒1/v = 1/60 ⇒ v = 60cm now, we should use formula of magnification m = v/u = height of image/height of object Here , v = 60cm, u = -30cm, height of object = 4cm , so, 60cm/-30cm = height of image/4cm-2 = height of image/4 height of image = -8cm , here negative sign shows image is formed below of optical axis { horizontal line }

Now, size of image/size of object = 8cm/4cm = 2 { excluding sign }

Hence, height of image or size of image = 8cm image distance form lens = 60cm , right side and ratio of image size or object size is 2 {excluding sign }

ray diagram of image , its position , principal focus are shown in figure.Where h shown height of image e.g., 8cm , v is shown distance of image from lens e.g., 60cm .


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