Question

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?

Answer 1

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

$\frac{\lambda (water)}{\lambda(air)}$ = $\frac{sin(\Theta)water}{sin(\Theta)air}$      

and

ratio of wavelength is = $\frac{1}{n}$

ratio of wavelength  = $\frac{1}{1.33}$

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 = $\frac{\Theta }{n}$

θ for water = $\frac{0.49}{1.33}$

θ for water = 0.3684°