Adouble-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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Ответ:
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
and
ratio of wavelength is =![\frac{1}{n}](/tpl/images/0417/5196/11e4c.png)
ratio of wavelength =![\frac{1}{1.33}](/tpl/images/0417/5196/a2a88.png)
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}](/tpl/images/0417/5196/fde48.png)
θ for water =![\frac{0.49}{1.33}](/tpl/images/0417/5196/709c2.png)
θ for water = 0.3684°
Ответ:
the displacement was 31
Explanation: