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`
(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 (iii) as said above.

Explanation

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,
`(1)/(v)-(1)/(u)=(1)/(f)`
`(1)/(f) =(1)/(38)-(1)/(-38)=(2)/(38)=(1)/(19)`
`f=19cm`
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 = -19 cm , 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