A mica strip and a polystyrene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0⋅50 mm and the separation between the slits is 0⋅12 cm. The refractive index of mica and polystyrene are 1.58 and 1.55, respectively, for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen at a distance one metre away. (a) What would be the fringe-width? (b) At what distance from the centre will the first maximum be located?
Answers
Given:
The thickness of the strips = t1=t2=t=0.5 mm=0.5×10⁻³m
Separation between the two slits d=0.12 cm=12×10⁻⁴m
refractive index of mica μm = 1.58
and of polystyrene, μp = 1.58
Wavelength of the light=λ=590 nm=590×10⁻⁹ m
Distance between screen and slit, D = 1 m
a)
fringe width is given by
β=λD/d
⇒β =590×10⁻⁹×1/12×10⁻⁴
=4.9×10⁻⁴ m
(b) When both the mica and polystyrene strips are fitted before the slits, the optical path changes by
∆x=(μm-1 )t- (μp-1) t =(μm-μp )t =(1.58-1.55)×0.5x 10⁻³
=0.015×10⁻³ m
∴ Number of fringes shifted, n = Δx/λ
=0.015×10⁻³/ 590×10⁻⁹=25.43
∴ 25 fringes and 0.43th of a fringe.
⇒ In which 13 bright fringes and 12 dark fringes and 0.43th of a dark fringe.
So, position of first maximum on both sides is given by
On one side,
x=0.43×4.91×10⁻⁴
∵β=4.91×10⁻⁴ m =0.021 cm
On the other side,
x’=(1-0.43)×4.91×10⁻⁴ =0.028 cm
Explanation:
Given: t
1
=t
2
=0.5mm=0.5×10
−3
m
μ
m
=1.58
μ
p
=1.55
λ=590nm=590×10
−9
m
d=0.12m=12×10
−4
m
D=1m
(a) Fringe width =
d
Dλ
=
12×10
−4
1×590×10
−9
=4.91×10
−4
m
(b) When both strips are fitted, the optical path changes by
Δx=(μ
m
−1)t
1
−(μ
p
−1)t
2
=(μ
m
−μ
p
)t
(1.58−1.55)×0.5×10
−3
=0.015×10
−3
m
So, Number of fringes shift =
590××10
−3
0.015×10
−3
=25.43
There are 25 fringes and 0.43
th
of a fringe.
There are 13 bright fringes and 12 dark fringes and 0.43
th
of a dark fringe. So, position of the first maximum on both sides will be given by
∴x=0.43×4.91×10
−4
=0.021cm
x
′
=(1−0.43)×4.91×10
−4
=0.028cm (since, fringe width=4.91×10
−4
m)