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CBSE Class 10 Science 2019 Delhi Set 1

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Question : 19 of 28
Marks: +1, -0
An object is placed at a distance of 60 cm60\ \mathrm{cm} from a concave lens of focal length 30 cm30\ \mathrm{cm}.
(i) Use lens formula to find the distance of the image from the lens.
(ii) List four characteristics of the image (nature, position, size, erect/inverted) formed by the lens in this case.
(iii) Draw ray diagram to justify your answer of part (ii).
Solution:  
(i)   1f  =  1v−  1u\;\frac{1}{f}\;=\;\frac{1}{v}-\;\frac{1}{u}  [lens formula]
  1f+  1u  =  1v\;\frac{1}{f}+\;\frac{1}{u}\;=\;\frac{1}{v}
u  =−60u\;=-60
f  =−30f\;=-30
(In case of Concave lens)
Hence, putting the values in formula, we get
  1−30+  1−60  =  1v\;\frac{1}{-30}+\;\frac{1}{-60}\;=\;\frac{1}{v}
  −2−160  =  1v\;\frac{-2-1}{60}\;=\;\frac{1}{v}
  −360  =  1v\;\frac{-3}{60}\;=\;\frac{1}{v}
  1v  =  −120\;\frac{1}{v}\;=\;\frac{-1}{20}
∴    v  =−20 cm\therefore \;\; v\;=-20\ \mathrm{cm}
∵    m  =  vu  =  −20−60=0.33\because \;\; m\;=\;\frac{v}{u}\;=\;\frac{-20}{-60}=0.33
(ii) So, the image formed will be virtual, erect, small in size and image will be formed 20 cm20\ \mathrm{cm} from lens on the same side as the object
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