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ICSE Class 10 Physics 2019 Paper

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An object is placed at a distance of 24 cm24\,\mathrm{cm} in front of a convex lens of focal length 8 cm8\,\mathrm{cm}.
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Question : 45 of 74
Marks: +1, -0
Calculate the distance of the image from the lens.
Solution:  
We know,
  1f  =  1v−  1u\; \frac{1}{f} \; = \; \frac{1}{v} - \; \frac{1}{u}
⇒      1v  =  1f+  1u\Rightarrow \;\; \; \frac{1}{v} \; = \; \frac{1}{f} + \; \frac{1}{u}
⇒      1v  =  18+  1−24\Rightarrow \;\; \; \frac{1}{v} \; = \; \frac{1}{8} + \; \frac{1}{-24}
  1v  =  18−  124\; \frac{1}{v} \; = \; \frac{1}{8} - \; \frac{1}{24}
  1v=  3−124=  224=  112\; \frac{1}{v} = \; \frac{3-1}{24} = \; \frac{2}{24} = \; \frac{1}{12}
or
v=12cmv = 12 \text{cm}
Thus the distance of the image from the lens is 12cm12 \text{cm} .
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