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Manipal Medical Manipal Medical Solved Paper-2003

  • question_answer
    A concave mirror of focal length 15 cm forms an image having twice the linear dimension of the object. The position of the object when the image is virtual will be:

    A)  45 cm         

    B)  30 cm

    C)  7.5 cm        

    D)  22.5 cm

    Correct Answer: C

    Solution :

     Focal length of concave mirror\[f=-15\text{ }cm\] magnification = 2 (for virtual image) Linear magnification \[m=\frac{size\text{ }of\text{ }image}{size\text{ }of\text{ }object}=-\frac{\upsilon }{u}\] \[2=-\frac{\upsilon }{u}\]or \[\upsilon =-2u\] Now from the relation \[\frac{1}{f}=\frac{1}{\upsilon }+\frac{1}{u}\] Or \[-\frac{1}{15}=\frac{1}{u}-\frac{1}{2u}=\frac{1}{2u}\] Or      \[2u=-15\] or \[n=-7.5\text{ }cm\]


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