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 [tex]z_1 - z_o = 1.44cm[/tex]
b
The distance of separation is [tex]z_4 - z_2 = 2.031cm[/tex]
Explanation:
From the question we are told that
The distance from the screen is [tex]D = 1.40m[/tex]
The slit separation is [tex]d = 0.0572 mm = 0.0572 *10^{-3} m[/tex]
The wavelength of the yellow light is [tex]\lambda_y = 598nm[/tex]
The distance of a fringe from the central maxima is mathematically represented as
[tex]z_n = n \frac{\lambda_y D}{d}[/tex]
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
[tex]z_1 - z_o = (1 - 0 ) \frac{\lambda_y D}{d}[/tex]
Substituting values
[tex]z_1 - z_o = (1 - 0 ) \frac{590*10^{-9} 1.40}{0.0572 *10^{-3}}[/tex]
[tex]z_1 - z_o = 0.0144m[/tex]
Converting to cm
[tex]z_1 - z_o = 0.0144m = 0.0144*100 = 1.44cm[/tex]
b
The wavelength of blue light is [tex]\lambda _b[/tex]
So the distance that separates second order from fourth order bright fringe can be evaluated as
[tex]z_4 - z_2 = (4 - 2 ) \frac{\lambda_y D}{d}[/tex]
Substituting values
[tex]z_4 - z_2 = (4 - 2 ) \frac{415*10^{-9} 1.40}{0.0572 *10^{-3}}[/tex]
[tex]z_4 - z_2 = 0.02031 \ m[/tex]
Converting to cm
[tex]z_4 - z_2 = 0.02031m = 0.02031*100 = 2.031cm[/tex]
