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CBSE Class 10 Science 2018 Solved Paper

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Question : 26 of 27
Marks: +1, -0
An object of height 4.0 cm4.0\ \mathrm{cm} is placed at a distance of 30 cm30\ \mathrm{cm} from optical centre ' OO ' of a convex lens of focal length 20 cm20\ \mathrm{cm}. Draw a ray diagram to find the position and size of the image formed. Mark optical centre '  O\ O ' and principal focus ' FF ' on the diagram. Also find the approximate ratio of size of image to the size of object.
Solution:  
Given, f=+20 cm, u=−30 cmf=+20\ \mathrm{cm},\ u=-30\ \mathrm{cm} h0=4 cmh_0=4\ \mathrm{cm}
We know that,
  1f=  1v−  1u\;\frac{1}{f}=\;\frac{1}{v}-\;\frac{1}{u}
  120=  1v−  1−30\;\frac{1}{20}=\;\frac{1}{v}-\;\frac{1}{-30}
  1v=  160\;\frac{1}{v}=\;\frac{1}{60}
⇒    v=60 cm\Rightarrow \;\; v=60\ \mathrm{cm}
  hih0=  vu\;\frac{h_i}{h_0}=\;\frac{v}{u}
  hi4=  60−30\;\frac{h_i}{4}=\;\frac{60}{-30}
So,⇒    hi=−8 cm\Rightarrow \;\; h_i=-8\ \mathrm{cm}
Thus, the height or size of the image is 8 cm8\ \mathrm{cm}. The minus sign shows that this height is in the downward direction, that is, the image is formed below the axis.
Ratio of size of image to object =−  84=−2=-\;\frac{8}{4}=-2
So image is enlarged beyond 2 F22\ \mathrm{F}_2.
Object between F1F_1 and 2 F12\ F_1.
Image is formed beyond 2 F22\ \mathrm{F}_2, real, inverted.
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