Question

In a double-slit interference experiment, the slit separation is 2.41 μm, the light wavelength is 512 nm, and the separation between the slits and the screen is 4.45 m. (a) What is the angle between the center and the third side bright fringe? If we decrease the light frequency to 94.5% of its initial value, (b) does the third side bright fringe move along the screen toward or away from the pattern's center and (c) how far does it move?

Answer 1

Answer:

Using equation 2dsinФ=n*λ

given d=2.41*10^-6m

λ=512*10^-12m

θ=52.64 degrees

Answer 2

Answer:

a

The angle between the center and the third side bright fringe  is

       $\theta = 39.60^o$

b

The third side bright fringe move away from the pattern's center

c

The distance by which it moves away is $\Delta z=0.3906 m$

Explanation:

From the question the

           The wavelength is $\lambda = 512nm$

In the first question we a asked to obtain the angle between the center and the third side bright fringe

  since we are considering the third side of the bright fringe the wavelength of light on  the three sides would be  evaluated as

                 $\lambda_{3} = 3 * 512nm$

The slit separation is given as  $d = 2.41 \mu m$

    The angle between the center and the third side bright fringe is

              $\theta = sin^{-1} (\frac{\lambda_3}{d} )$

              $\theta = sin^{-1} (\frac{3 *512*10^{-9}}{2.24*10^{-6}} )$

                $= sin^{-1} (0.6374)$

                $\theta = 39.60^o$

When the frequency of the light is reduced the wavelength is increased

                i.e $f = \frac{c}{\lambda}$

and this increase would cause the third side bright to move away from the pattern's center

        Now from the question frequency is reduce to 94.5% this mean that the  wavelength would also increase by the same as mathematically represented below

                  $\lambda_{new} = \frac{512 *10^{-9}}{0.945}$

                          $= 0.542 \mu m$

The angle between the center and the third side bright fringe is for new wavelength

                   $\theta = sin^{-1} (\frac{3 *512*10^{-9}}{2.41*10^{-6}} )$

                     $= 42.46^o$

 The distance traveled away from the pattern's center is mathematically represented as

                $z = A tan \theta$

 Where A is the separation between the slits and the screen

              $\Delta z = 4.45(tan 42.46 - tan39.60 )$

                    $\Delta z=0.3906 m$