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CBSE Class 10 Science 2016 Term II Delhi Set 1

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Question : 23 of 36
Marks: +1, -0
(a) Draw a ray diagram to ss formation of image by a convex lens when an object is placed in front of the lens between its optical centre and principal focus.
(b) In the above ray diagram mark the objectdistance (u)(u) and the image-distance (v)(v) with their proper signs (+ve or - ve as per the new Cartesian sign convention) and state how these distances are related to the focal length (f)(f) of the convex lens in this case.
(c) Find the power of a convex lens which forms a real, and inverted image of magnification -1 of an object placed at a distance of 20 cm20\ \mathrm{cm} from its optical centre.
Solution:  
(b) The object distance and image distance are marked in the diagram of part (a).
The relation between v,uv, u and ff is given by the formula,
  1f=  1v−  1u\;\frac{1}{f}=\;\frac{1}{v}-\;\frac{1}{u}
(c) Given, m=−1 ; u=−20 cm ; v=m=-1\ ;\ u=-20\ \mathrm{cm}\ ;\ v= ?, f=f= ?
We know that, m=  vum=\;\frac{v}{u}
−1    =    v−20-1\;\;=\;\;\frac{v}{-20}
∴     v  =+20 cm\therefore\ \;\;v\;=+20\ \mathrm{cm}
Thus, object is at 2F2F.
   i.e.,     2f=20 cm\;\text{ i.e., }\;\;2 f=20\ \mathrm{cm}
∴  f=  202=10 cm=0.1 m\therefore \;f=\;\frac{20}{2}=10\ \mathrm{cm}=0.1\ \mathrm{m}
  P=  1f=  10.1=+10 D\; P=\;\frac{1}{f}=\;\frac{1}{0.1}=+10\ \mathrm{D}
The power of convex lens is +10 D+10\ \mathrm{D}.
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