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CBSE Class 10 Science 2015 Term II Delhi Set 2

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Question : 11 of 11
Marks: +1, -0
The image of a candle flame placed at a distance of 30 cm30\,\text{cm} from a spherical lens is formed on a screen placed on the other side of the lens at a distance of 60 cm60\,\text{cm} from the optical centre of the lens. Identify the type of lens and calculate its focal length. If the height of the flame is 3 cm3\,\text{cm}, find the height of its image.
Solution:  
Given:
h1=+3 cm,u=−30 cmh_1=+3\,\text{cm}, u=-30\,\text{cm}
v=+60 cm,f=   ?,   h′=   ?   v=+60\,\text{cm}, f=\;\text{ ?, }\; h'=\;\text{ ? }\;
  1f=  1v−  1u=  160−  1−30\;\frac{1}{f}=\;\frac{1}{v}-\;\frac{1}{u}=\;\frac{1}{60}-\;\frac{1}{-30}
  1f=  +360\;\frac{1}{f}=\;\frac{+3}{60}
∴    f=+20 cm\therefore\;\; f=+20\,\text{cm}
$ $
Therefore, the lens is convex.
h′  =  vu×h  =  +60 cm−30 cm×3 cmh'\;=\;\frac{v}{u} \times h\;=\;\frac{+60\,\text{cm}}{-30\,\text{cm}} \times 3\,\text{cm}
h′  =−6 cmh'\;=-6\,\text{cm}
Here, negative sign indicates that the image formed is inverted. Therefore, height of image of candle flame is 6 cm6\,\text{cm}.
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