Answer to Question #95991 in Optics for Subia Gulzar

Question

A diffraction phenomenon is observed using a double slit illuminated with light of wavelength 5000 °A. The slit with is 0.02mm and spacing between the two slit is 0.1mm. The distance of the screen from the slits is 100cm. Calculate
(1) distance between central maximum and first minimum of the fringes envelope.
(2) distance between any two consecutive double slit dark fringes
(3)amplitude and intensity.

Answer 1

The angular separation between central maximum and first minimum of the fringes envelope is given by formulas

$\sin {θ_1}= θ_1=\frac{λ}{a+b} (1)$

$θ_1=\frac{x_1}{d} (2)$

Using (1) and (2) we have

$\frac{x_1}{d}=\frac{λ}{a+b} (3)$

Using (3) we get

$x_1=\frac{λd}{2(a+b)} (4)$

In our case, a=2×10^-5 m, b=10^-4 m, λ=5×10^-7 m, d=1 m

We get

x₁=2.08 mm

The angular separation between two consecutive double slit dark fringes is given by formula

$\sin {θ_1}-\sin {θ_2}= θ_1- θ_2=\frac{λ}{2(a+b)} (5)$

$θ_1- θ_2=\frac{x_2}{d} (6)$

Using (5) and (6) we get

x₂= 4.16 mm