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

Let f be the focal length of the convex lens. The distance of object should be measured from pole of the lens.
Distance of candle (or object)
= Position of convex lens-Position of candle
= 50 - 12 = 38 cm.
Now, by sign convention, distance of candle (or object) = u = -38 cm
Similarly, distance of candle.s image = position of the screen -position of convex lens
= 88 - 50 = 38 cm
By sign convention, distance of candle.s image = v = + 38 cm
(i) Using lens formula,
`1/v-1/u=1/f`
`1/f=(1)/(38) -(3)/(-38)=(1)/(19) implies f = 19 cm`
The focal length of the convex lens is 19 cm.
(ii) When the candle is shifted towards the lens at a position of 31.0 cm. Then, new object distance = position of convex lens - position of candle = 50 - 31 = 19
By sign convention, u = -19 cm.
Now, focal length of the convex lens = 19 cm. It means, the candle lays at the focus of lens, hence its image is formed at infinity.
(iii) When he further shifts the candle towards the lens. This means candle lies between optical centre and focus of convex lens, so, magnified, virtual and erect image of the candle will be formed.
(iv) The ray diagram of image formation is given below: