question_answer 69)

  A student focused the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle, screen and the lens as under position of candle =12.0 cm position of convex lens = 50.0 cm position of the screen = 88.0 cm (i) What is the focal length of the convex lens? (ii) Where will the image be formed, if he shifts the candle towards the lens at a position of 31.0 cm? (iii) What will be the nature of the image formed, if he further shifts the candle towards the lens? (iv) Draw a ray diagram to show the formation of the image in case as said above.                    

Answer:

                  Object distance u = Position of the convex lens - Position of candle                                 =50 ? 12=38 cm By sign convention ,u = -38cm Similarly, image distance, v = position of the screen - position of convex lens                 = 88 ? 50 = 38 cm By sign convention,         v = + 38 cm                 (i) Using lens formula, cm The focal length of the convex lens is 19 cm. (ii) After shifting the candle towards the lens at a position of 31.0 cm, then object distance u = position of convex lens - position of candle                = 50 ? 31 =19 By sign convention, u = - 19cm, but here focal length of the convex lens is also 19 cm. So, the candle lies at the focus of lens, hence its image forms at infinity. (iii) When student further shifts the candle towards the lens i.e., candle lies between optical centre and focus of convex lens, then lens forms enlarged, virtual and erect image of the candle. (iv) The ray diagram showing the formation of image