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`
(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.

Answer 1

Here, `u =` distance of candle flame from the lens
`= 50.0 - 12.0 = 38 cm = -38 cm` (as per New Cartesian Sign Convention)
`v =` distance of screen from the lens
`= 88.0 - 50.0 = 38 cm` (as per new Cartesian Sign Convention)
(i) From `(1)/(f) = (1)/(v) - 1/u`
`(1)/(f) = 1/38 + 1/38 = 2/38, f = 38/2 = 19 cm`.
(ii) When the student shifts the candle towards the lens at a position of `31.0 cm`,
` u = 50.0 - 31.0 = 19 cm`.
The candle lies at the focus of convex lens.
The image will be formed at infinity.
(iii) When he further shifts the candle towards the lens, the candle lies between optical centre of the lens and its focus. Therefore, the image formed is virtual, erect and magnified.
(iv) For ray diagram in case (iii) above