12th Class Physics Sample Paper Physics Sample Paper-9

  • question_answer
    A man with normal near point (25 cm) reads a book with small print using a magnifying glass of a thin convex lens of focal length 5 cm,
    (i) What is the closest and the farthest distance at which he should keep the lens from the page, so that he can read the book when viewing through the magnifying glass?
    (ii) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?


    (i) Given, focal length of convex lens, \[f=+5cm.\] For closest distance, \[v=-25\,cm\] (normal near point for distinct vision) \[\therefore \] From lens formula, \[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\] \[\Rightarrow \]   \[\frac{1}{{{u}_{\min }}}=\frac{1}{v}-\frac{1}{f}=-\frac{1}{25}-\frac{1}{5}=\frac{-1-5}{25}\] \[\Rightarrow \]   \[{{u}_{\min }}=-\frac{25}{6}=-4.2\,cm\] For farthest distance, \[v=\infty \] \[\frac{1}{{{u}_{\max }}}=\frac{1}{v}-\frac{1}{f}=\frac{1}{\infty }-\frac{1}{5}\]           [by lens formula] \[\Rightarrow \]   \[{{u}_{\max }}=-5cm\] Thus, the nearest distance is 4.2 cm and the farthest distance is 5 cm (focal length of lens) for reading a book. (ii) Angular magnification,            \[M=\frac{D}{u}\] and distance of distinct vision, \[D=-25\,cm\] Maximum angular magnification,             \[{{M}_{\max }}=\frac{(-25)}{{{u}_{\min }}}=\frac{(-25)}{(-25/6)}=6\] Minimum angular magnification,             \[{{M}_{\min }}=\frac{(-25)}{{{u}_{\max }}}=-\frac{25}{(-5)}=+5\]

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