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NEET-XII-Physics

21: Speed of Light

with Solutions - page 2

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    Section : iv
  • Qstn #1
    In an experiment to measure the speed of light by Fizeau’s apparatus, following data are used: Distance between the mirrors = 12.0 km,
    Number of teeth in the wheel = 180.
    Find the minimum angular speed of the wheel for which the image is not seen.
    Ans : Distance between the mirrors (D) = 12.0 km = 12 × 103 m
    Number of teeth in the wheel (n) = 180
    Now we apply Fizeau's apparatus
    Speed of light, c = 3 × 108 m/s
    `` \,\mathrm{\,We\,}\,\mathrm{\,know\,},c=\frac{2Dn\omega }{\,\mathrm{\,\pi \,}}``
    `` \Rightarrow \omega =\frac{c\pi }{2Dn}\,\mathrm{\,red\,}/\,\mathrm{\,s\,}``
    `` =\frac{\pi c}{2\,\mathrm{\,D\,}n}\times \frac{180}{\pi }\,\mathrm{\,deg\,}/\,\mathrm{\,s\,}``
    `` \omega =\frac{3\times {10}^{8}}{24\times {10}^{3}}``
    `` =1.25\times {10}^{4}\,\mathrm{\,deg\,}/\,\mathrm{\,sec\,}``
    Hence, the required angular speed of the wheel for which the image is not seen is `` 1.25\times {10}^{4}\,\mathrm{\,deg\,}/\,\mathrm{\,sec\,}``
    Page No 448:
  • Qstn #2
    In an experiment with Foucault’s apparatus, the various distances used are as follows:
    Distance between the rotating and the fixed mirror = 16 m
    Distance between the lens and the rotating mirror = 6 m,
    Distance between the source and the lens = 2 m.
    When the mirror is rotated at a speed of 356 revolutions per second, the image shifts by 0.7 mm. Calculate the speed of light from these data.
    Ans : Distance between the rotating and the fixed mirror (R) = 16 m
    Distance between the lens and the rotating mirror (b) = 6 m
    Distance between the source and the lens (a) = 2 m
    Mirror is rotated at a speed of 356 revolutions per second
    ⇒ ω = 356 rev/s= 356 × 2 π rad/sec
    Shift in the image (s) = 0.7 m = 0.7 × 103 m
    In Foucault experiment, speed of light is given by
    `` c=4{\,\mathrm{\,R\,}}^{2}\frac{wa}{s(\,\mathrm{\,R\,}+b)}``
    `` =\frac{4\times (16{)}^{2}\times 356\times 2\,\mathrm{\,\pi \,}\times 2}{(0.7)\times {10}^{-3}(16+6)}``
    = 2.975 × 108 m/s
    Therefore, the required speed of light is 2.975 × 108 m/s.
    Page No 448:
  • Qstn #3
    In a Michelson experiment for measuring speed of light, the distance travelled by light between two reflections from the rotating mirror is 4.8 km. The rotating mirror has a shape of a regular octagon. At what minimum angular speed of the mirror (other than zero) the image is formed at the position where a nonrotating mirror forms it?
    Ans : Distance travelled by light between two reflections from the rotating mirror (D) = 4.8 km = 4.8 × 103 m
    Number of faces of the mirror, N = 8
    Angular speed of the mirror, `` \omega `` = ?
    In Michelson experiment, the speed of light (c) is given by
    `` c=\frac{\omega DN}{2\,\mathrm{\,\pi \,}}``
    where
    ω = angular speed
    N = number of faces in the polygon mirror
    `` \therefore \omega =\frac{2\,\mathrm{\,\pi \,}c}{DN}rad/\,\mathrm{\,s\,}``
    `` =\frac{c}{DN}\,\mathrm{\,rev\,}/\,\mathrm{\,s\,}``
    `` =\frac{3\times {10}^{8}}{(4.8\times {10}^{3}\times 8)}``
    = 7.8 × 103 rev/s
    Hence, the required angular speed is 7.8 × 103 rev/s.
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