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VMMC VMMC Medical Solved Paper-2002

  • question_answer
    If an object is placed at 10 cm in front of a concave mirror of focal length 15 cm. The magnification of image is:

    A)  -1.5                                      

    B)  1.5                        

    C)         -3                          

    D)         3

    Correct Answer: C

    Solution :

    Now from the formula for mirror \[\frac{1}{\upsilon }+\frac{1}{u}=\frac{1}{f}\]                                     ?(1) Multiplying Eq. (1) by u on both sides \[\frac{u}{\upsilon }+1=\frac{u}{f}\]or \[\frac{1}{m}+1=\frac{u}{f}\]                        (As \[\frac{\upsilon }{u}=m\]) or            \[\frac{1}{m}=\frac{u}{f}-1\]or \[m=\frac{f}{u-f}\]                           ?(2) Putting the given values in Eq. (2), we obtain \[m=\frac{-15}{-10+15}=-\frac{15}{5}=-3\]                 Hence,                  \[m=-3\]


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