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CBSE Class 10 Science 2021 Term 1 Solved Paper

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Question : 43 of 60
Marks: +1, -0
3. A converging lens forms a three times magnified image of an object, which can be take on a screen. If the focal length of the lens is 30 cm30\,\text{cm}, then the distance of the object from the lens is:
Solution:  
Explanation: We know f=R2f=\frac{R}{2}
  m=vu    (m= magnification,   \; m = \frac{v}{u} \;\; ( m = \text{ magnification, } \;
  v= image distance,   u= object distance)   \; v = \text{ image distance, } \; u = \text{ object distance) } \;
    −3u=v\;\; -3u = v
  1v−1u=1f\; \frac{1}{v} - \frac{1}{u} = \frac{1}{f}
    −13u−1u=130\;\; -\frac{1}{3u} - \frac{1}{u} = \frac{1}{30}
    −u−3u3u=130\;\; -u - \frac{3u}{3u} = \frac{1}{30}
    −4u3u2=130\;\; -\frac{4u}{3u^2} = \frac{1}{30}
    −3u4=30\;\; -\frac{3u}{4} = 30
  u=−40 cm\; u = -40\,\text{cm}
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