Question

Two slits separated by a distance of d = 0.12 mm are located at a distance of D = 0.63 m from a screen. The screen is oriented parallel to the plane of the slits. The slits are illuminated by a coherent light source with a wavelength of λ = 540 nm. A wave from each slit propagates to the screen. The interference pattern shows a peak at the center of the screen (m=0) and then alternating minima and maxima. What is the pathlength difference between the waves at the second maximum (m=2) on the screen?

Answer 1

Answer:

The path-length difference is $dsin\theta=1.08*10^{-3}mm$

The angle is  $\theta = 0.5157^o$

Explanation:

  From the question we are told that

             The distance of separation is  d = 0.12 mm = $0.12*10^{-3} m$

             The distance from the screen is  D = 0.63 m

              The wavelength is $\lambda = 540nm = 540 *10^{-9}m$

From the question we can deduce that the the two  maxima's are at the m=0 and m=2

   Now the path difference for this second maxima is mathematically represented as

                   $d sin \theta = m \lambda$

  Where d$dsin\theta$ is the path length difference

Substituting values

        $dsin \theta = 2 * 540*10^{-9}$

                 $dsin\theta = 1.08*10^{-6}m$

converting to mm

               $dsin\theta = 1.08*10^{-6} * 1000 mm$

                        $dsin\theta=1.08*10^{-3}mm$

To obtain the angle we make $\theta$ the subject

             $\theta = sin ^{-1} [\frac{m \lambda}{d} ]$

 Substituting values

             $\theta = sin ^{-1} [\frac{1.08*10^{-6}}{0.12*10^-3} ]$

               $\theta = 0.5157^o$