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JEE Main & Advanced AIEEE Solved Paper-2012

  • question_answer
    An object 2.4 m in front of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1 cm thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object shifted to be in sharp focus on film?   AIEEE  Solved  Paper-2012

    A) 7.2 m

    B) 2.4 m

    C) 3.2 m                                       

    D) 5.6 m

    Correct Answer: D

    Solution :

                 \[\frac{1}{f}=\frac{1}{12}+\frac{1}{240}=\frac{20+1}{240}\]              \[f=\frac{240}{21}m\] shift \[=1(1-\frac{2}{3})=\frac{1}{3}\] Now v? \[=12-\frac{1}{3}=\frac{35}{3}\,\,cm\] \[\therefore \frac{21}{240}=\frac{3}{35}-\frac{1}{u}\] \[\frac{1}{u}=\frac{3}{35}-\frac{21}{240}=\frac{1}{5}\left( \frac{3}{7}-\frac{21}{48} \right)\] \[\frac{5}{u}=\left| \frac{144-147}{48\times 7} \right|\] \[u=560\,\,cm=5.6\,m\]


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