Light of wavelength 710 nm passes through two narrow slits 0.66 mm apart. The screen is 2.00 m away. A second source of unknown wavelength p

Question

Light of wavelength 710 nm passes through two narrow slits 0.66 mm apart. The screen is 2.00 m away. A second source of unknown wavelength produces its second-order fringe 1.25 mm closer to the central maximum than the 710-nm light. What is the wavelength of unknown light?

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Hồng Cúc 6 months 2021-09-04T19:58:12+00:00 1 Answers 0 views 0

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    2021-09-04T20:00:04+00:00

    Answer:

    The wavelength is 503 nm  

    Explanation:

    Considering constructive interference , this means that route(path) difference is equal to the product of order of fringe and wavelength of the light

     i.e   dsinθ  = m\lambda

    Where \lambda  is the wavelength of light  and m is the order of the fringe

        Looking at θ to be very small , sin θ can be approximated to  θ

                  and   \theta \approx \frac{x}{l}

    Substituting this into the above equation

               d[\frac{x}{l} ] =m\lambda

          making x the subject

                        x =\frac{m\lambda l}{d}

          This above equation will give the value of the distance of the m^{th} order fringe of the wavelength \lambda from the central fringe

           Replacing with the value given in the question we have

      \lambda = 710 nm  m = 2  d =0.66 mm , l = 2.0 m

                      x = \frac{(2)(710nm)(2.0m)[\frac{10^9}{1m} ]}{(0.66mm)(\frac{10^6}{1mm} )}

                       =(4.303*10^6nm)[\frac{\frac{1}{10^6}mm }{1nm} ]

                        =4.303mm

     The separation of the second fringe from central maximum is 4,303 mm    

       

    To obtain the separation of the second order fringe of the unknown light from central maximum

           x' = 4.303mm - 1.25 mm = 3.053mm

    Now to obtain the wavelength of this second source

                       from x = \frac{m\lambda l}{d}

                           \lambda' = \frac{x'd}{ml}

    Now substituting 3,053 mm for x' 2.0 mm for l , 0.66 mm  for d and 2 for m in the above formula

               \lambda' =\frac{(3.053mm)(0.66mm)}{(2)(2.0)(\frac{10^3mm}{1m} )}

                      = (503.7*10^{-6}mm)(\frac{10^6nm}{1mm} )

                       =503.7nm

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