Question

A student locates a double-slit assembly 1.40 m from a reflective screen. The slits are separated by 0.0572 mm. (a) Suppose the student aims a beam of yellow light, with a wavelength of 589 nm, toward the slit assembly, and this makes an interference pattern on the screen. What distance (in cm) separates the zeroth-order and first-order bright fringes (a.k.a. maxima)?

Answer 1

Complete Question

A student locates a double-slit assembly 1.40 m from a reflective screen. The slits are separated by 0.0572 mm.

(a) Suppose the student aims a beam of yellow light, with a wavelength of 589 nm, toward the slit assembly, and this makes an interference pattern on the screen. What distance (in cm) separates the zeroth-order and first-order bright fringes (a.k.a. maxima)?

(b)

Now suppose that blue light (with

λ = 415 nm)

is used instead. What distance (in cm) will now separate the second-order and fourth-order bright fringes?

Answer:

a

The distance of separation is  $z_1 - z_o = 1.44cm$

b

The distance of separation is  $z_4 - z_2 = 2.031cm$

Explanation:

From the question we are told that

     The distance from the screen is  $D = 1.40m$

      The slit separation is $d = 0.0572 mm = 0.0572 *10^{-3} m$

       The wavelength of the yellow light is  $\lambda_y = 598nm$

       

The distance of a fringe from the central maxima is mathematically represented as

       $z_n = n \frac{\lambda_y D}{d}$

Where n is the order of the fringe so the distance of separation between  

   The distance that separates first order from zeroth order bright fringe can be evaluated as

            $z_1 - z_o = (1 - 0 ) \frac{\lambda_y D}{d}$

Substituting values

          $z_1 - z_o = (1 - 0 ) \frac{590*10^{-9} 1.40}{0.0572 *10^{-3}}$

          $z_1 - z_o = 0.0144m$

Converting to cm

            $z_1 - z_o = 0.0144m = 0.0144*100 = 1.44cm$

b

The  wavelength of blue light is  $\lambda _b$

       So the distance that separates second  order from fourth order bright fringe can be evaluated as

         $z_4 - z_2 = (4 - 2 ) \frac{\lambda_y D}{d}$

Substituting values

        $z_4 - z_2 = (4 - 2 ) \frac{415*10^{-9} 1.40}{0.0572 *10^{-3}}$

        $z_4 - z_2 = 0.02031 \ m$

Converting to cm

            $z_4 - z_2 = 0.02031m = 0.02031*100 = 2.031cm$