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CBSE Class 10 Science 2021 Term 1 Solved Paper

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Question : 47 of 60
Marks: +1, -0
An object of height 4 cm4\text{ cm} is kept at a distance of 30 cm30\text{ cm} from the pole diverging mirror. If the focal length of the mirror is 10 cm10\text{ cm}, the height of the image formed is
Solution:  
Explanation: Height of the object h=4 cmh=4\text{ cm}
Object distance, u=−30 cmu=-30\text{ cm}
Focal length, f=10 cmf=10\text{ cm} (Diverging mirror)
Using mirror formula,
1v+1u  =1f\frac{1}{v}+\frac{1}{u}\;=\frac{1}{f}
uv  =uf−1\frac{u}{v}\;=\frac{u}{f}-1
  =−3010−1\;=\frac{-30}{10}-1
  =4\;=4
Magnification, m=m= image height/Object height =−vu=-\frac{v}{u}
=14=\frac{1}{4}
Image height =h4=44=1 cm=\frac{h}{4}=\frac{4}{4}=1\text{ cm}
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