A double-slit arrangement produces bright interference fringes for sodium light ( λ = 603 nm) that are angularly separated by 0.49° near the center of the pattern. What is the angular fringe separation if the entire arrangement is immersed in water, which has an index of refraction of 1.33?

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DeniceSandidge DeniceSandidge · Answer 1 · 2019-12-15T07:55:54+01:00

Answer:

θ for water = 0.3684°

Explanation:

given data

wavelength λ = 603 nm

angle θ = 0.49°

index of refraction n = 1.33

to find out

angular fringe separation

solution

we know that double slit interference is here

m λ = d sin(θ) .................1

so for air it will be

m λ(air) = d sin(θ)air ...........2

and for water it will be

m λ(water) = d sin(θ)water .............3

now we take here ratio of equation 2 and 3

[tex]\frac{\lambda (water)}{\lambda(air)}[/tex] = [tex]\frac{sin(\Theta)water}{sin(\Theta)air}[/tex]

and

ratio of wavelength is = [tex]\frac{1}{n}[/tex]

ratio of wavelength = [tex]\frac{1}{1.33}[/tex]

ratio of wavelength = 0.75187

so

sin(θ)water = sin(θ)air (0.75187)

sin(θ)water = sin(0.49) (0.75187)

sin(θ)water = 0.006429

so (θ)water is

(θ)water = 0.3683°

we notice here that by using small angle formula

we have approximate sin(θ) = θ

so θ for water is = [tex]\frac{\Theta }{n}[/tex]

θ for water = [tex]\frac{0.49}{1.33}[/tex]

θ for water = 0.3684°