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CBSE Class 10 Science 2017 Delhi Term II Set 2 Paper

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Question : 9 of 9
Marks: +1, -0
A 3 cm3\ \text{cm} tall object is placed 18 cm18\ \text{cm} in front of a concave mirror of focal length 12 cm12\ \text{cm}. At what distance from the mirror should a screen be placed to see a sharp image of the object on the screen. Also calculate the height of the image formed.
Solution:  
Using mirror equation :
u=−18 cm, f=−12 cm, v=?, h1=+3 cm, h2=?u=-18\ \text{cm},\ f=-12\ \text{cm},\ v=?,\ h_1=+3\ \text{cm},\ h_2=?
  1v+  1u=  1f\;\frac{1}{v}+\;\frac{1}{u}=\;\frac{1}{f}
⇒      1v−  118  =−  112\Rightarrow \;\; \;\frac{1}{v}-\;\frac{1}{18}\;=-\;\frac{1}{12}
  1v  =−  112+  118\;\frac{1}{v}\;=-\;\frac{1}{12}+\;\frac{1}{18}
⇒      1v  =  −3+236=  −136\Rightarrow \;\; \;\frac{1}{v}\;=\;\frac{-3+2}{36}=\;\frac{-1}{36}
v=−36 cmv=-36\ \text{cm}
Thus, to obtain a sharp image of the object, the screen should be placed at a distance of 36 cm36\ \text{cm}, in front of the mirror.
Now,
m=  −vu=  h2h1m=\;\frac{-v}{u}=\;\frac{h_2}{h_1}
m=−(  −36−18)=−2m=-\left(\;\frac{-36}{-18}\right)=-2
or
−2  =  h23-2\;=\;\frac{h_2}{3}
h2=−6 cmh_2=-6\ \text{cm}
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