In a two-slit experiment, the slit separation is 3.34 ⋅ 10 − 5 m. The interference pattern is created on a screen that is 3.30 m away from t

Question

In a two-slit experiment, the slit separation is 3.34 ⋅ 10 − 5 m. The interference pattern is created on a screen that is 3.30 m away from the slits. If the 7th bright fringe on the screen is 29.0 cm away from the central fringe, what is the wavelength of the light?

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Thiên Di 3 years 2021-08-09T16:10:52+00:00 1 Answers 7 views 0

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    2021-08-09T16:12:21+00:00

    Answer:

    The wavelength is  \lambda  = 419 \ nm

    Explanation:

    From the question we are told that

       The  distance of separation is   d =  3.34 *10^{-5} \ m

       The  distance of the screen is  D  = 3.30 \ m

          The  order of the fringe is  n =  7

         The distance of separation of  fringes is y =  29.0 cm = 0.29 m

       

    Generally the wavelength of the light is mathematically represented as

              \lambda  =  \frac{y *  d }{ n  *  D}

    substituting values

             \lambda  =  \frac{0.29  *  3.34*10^{-5} }{ 7  *  3.30}

            \lambda  = 4.19*10^{-7}\ m

            \lambda  = 419 \ nm

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