Question

(a) How many fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern if λ = 471 nm, d = 0.117 mm, and a = 35.7 µm? (b) What is the ratio of the intensity of the third bright fringe to the intensity of the central fringe?

Answer 1

Answer:

a

The number of fringe is  z  = 3 fringes

b

The  ratio is $I = 0.2545I_o$

Explanation:

a

 From the question we are told that

        The wavelength is  $\lambda = 600 nm$

        The distance between the slit is  $d = 0.117mm = 0.117 *10^{-3} m$

        The width of the slit is  $a = 35.7 \mu m = 35.7 *10^{-6}m$

let  z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is  and this mathematically represented as

             $z = \frac{d}{a}$

Substituting values

             $z = \frac{0.117*10^{-3}}{35.7 *10^{-6}}$  

             z  = 3 fringes

b

   From the question  we are told that the order  of the bright fringe is  n = 3

   Generally the intensity of  a pattern  is mathematically represented as

                 $I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]$

Where $I_o$ is the intensity  of the  central fringe

 And  Generally  $sin \theta = \frac{n \lambda }{d}$

               $I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]$

               $I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]$

               $I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]$

                $I = I_o (1)(0.2545)$

                  $I = 0.2545I_o$