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CBSE Class 10 Science 2015 Term II Outside Delhi Set 3

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Question : 10 of 10
Marks: +1, -0
An object of height 5 cm5\ \mathrm{cm} is placed perpendicular to the principal axis of a concave lens of focal length 10 cm10\ \mathrm{cm}. If the distance of the object from the optical centre of the lens is 20 cm20\ \mathrm{cm}, determine the position, nature and size of the image formed using the lens formula.
Solution:  
Given : h1=+5 cm,f=−10 cm,u=−20 cmh_1=+5\ \mathrm{cm}, f=-10\ \mathrm{cm}, u=-20\ \mathrm{cm}
We know that,
  1f=  1v−  1u\;\frac{1}{f}=\;\frac{1}{v}-\;\frac{1}{u}
  1v=  1f+  1u=  1−10      120=  −320\;\frac{1}{v}=\;\frac{1}{f}+\;\frac{1}{u}=\;\frac{1}{-10} \;\; \;\frac{1}{20}=\;\frac{-3}{20}
or
Image distance, v=−  203 cmv=-\;\frac{20}{3}\ \mathrm{cm}.
The nature of the image is virtual and erect.
Now, magnification, m=  h2h1=  vum=\;\frac{h_2}{h_1}=\;\frac{v}{u}
⇒h2=  vu×h1=  −203×  1−20×5=  +53 cm\Rightarrow h_2=\;\frac{v}{u} \times h_1=\;\frac{-20}{3} \times \;\frac{1}{-20} \times 5=\;\frac{+5}{3}\ \mathrm{cm}
∴    \therefore \;\; The size of the image is 1.67 cm1.67\ \mathrm{cm}.
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