(a) How many fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern if

Question

(a) How many fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern if λ = 471 nm, d = 0.117 mm, and a = 35.7 µm? (b) What is the ratio of the intensity of the third bright fringe to the intensity of the central fringe?

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Kiệt Gia 6 months 2021-07-29T20:19:32+00:00 1 Answers 10 views 0

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  1. King
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    2021-07-29T20:21:19+00:00
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    Answer:

    a

    The number of fringe is  z  = 3 fringes

    b

    The  ratio is I = 0.2545I_o

    Explanation:

    a

     From the question we are told that

            The wavelength is  \lambda = 600 nm

            The distance between the slit is  d = 0.117mm = 0.117 *10^{-3} m

            The width of the slit is  a = 35.7 \mu m = 35.7 *10^{-6}m

    let  z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is  and this mathematically represented as

                 z = \frac{d}{a}

    Substituting values

                 z = \frac{0.117*10^{-3}}{35.7 *10^{-6}}  

                 z  = 3 fringes

    b

       From the question  we are told that the order  of the bright fringe is  n = 3

       Generally the intensity of  a pattern  is mathematically represented as

                     I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]

    Where I_o is the intensity  of the  central fringe

     And  Generally  sin \theta = \frac{n \lambda }{d}

                   I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]

                   I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]

                   I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]

                    I = I_o (1)(0.2545)

                      I = 0.2545I_o

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