A red laser from the physics lab is marked as producing 632.8-nm light. When light from this laser falls on two closely spaced slits, an int

Question

A red laser from the physics lab is marked as producing 632.8-nm light. When light from this laser falls on two closely spaced slits, an interference pattern formed on a wall several meters away has bright red fringes spaced 6.00 mm apart near the center of the pattern. When the laser is replaced by a small laser pointer, the fringes are 6.19 mm apart. What is the wavelength of light produced by the pointer?

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Nick 5 years 2021-08-06T12:04:27+00:00 1 Answers 18 views 0

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  1. sori
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    2021-08-06T12:05:44+00:00
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    Answer:

    The wavelength is  \lambda_R = 649 *10^{-9}\ m

    Explanation:

    From the question we are told that

       The wavelength of the red laser is  \lambda_r = 632.8 \ nm = 632.8 *10^{-9}\ m

        The spacing between  the fringe is  y_r = 6.00\ mm = 6.00*10^{-3} \ m

       The spacing between  the fringe for smaller laser point  is  y_R = 6.19 \ mm = 6.19 *10^{-3} \ m

          Generally the spacing between  the fringe is mathematically represented as

           y = \frac{D * \lambda }{d}

    Here  D is the distance to the screen

        and  d is the distance of the slit separation

    Now for both laser red light light and  small laser  point  D and  d are same for this experiment

    So

             \frac{y_r}{\lambda_r} = \frac{D}{d}

    =>      \frac{y_r}{\lambda_r} = \frac{y_R}{\lambda_R}

    Where \lambda_R  is the wavelength produced by the small laser pointer

      So

               \frac{6.0 *10^{-3}}{ 632.8*10^{-9}} = \frac{ 6.15 *10^{-9}}{\lambda_R}

    =>       \lambda_R = 649 *10^{-9}\ m

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