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?

Answer 1

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

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

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

                   $=503.7nm$