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