Question

A student focussed 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

1.  What is the focal length of the convex lens?
2. Where will the image be formed if he shifts the candle towards the lens at a position of 31.0 cm?
3. What will be the nature of the image formed if he further shifts the candle towards the lens?
4. Draw a ray diagram to show the formation of the image in case (iii) as said above.

Answer 1

Given:
Position of candle = 12.0 cm,

Position of the screen = 88.0 cm,

Position of convex lens = 50.0 cm

Position of object (u) = - (50 - 12) = -38 cm,

Position of image (υ) = 88 - 50 = 38 cm

1. Using the lens formula,
   `1/upsilon - 1/"u" = 1/"f"`
   `1/38 - 1/(-38) = 1/"f"`
   `2/38 = 1/"f"`
   f = 19 cm
   The focal length of the convex lens is 19 cm.
   
   
2. Given: 
   Position of object (u) = - (50 - 31) = -19 cm
   Using the lens formula,
   `1/upsilon - 1/"u" = 1/"f"`
   `1/upsilon - 1/(-19) = 1/19`
   `1/upsilon = 0`
   υ = infinity
   
   
3. In the previous case, the candle was kept at the focus, if he further shifts the candle towards the lens. The virtual, erect and magnified image will be formed.
4. 
   
   
5.