Question #93248

In a double slit experiment, the wavelength of the light source is 405nm, the slit separation d is 19.44*10^-6m and the slits width a is 4.050*10^-6m. Consider the interference of the light from the two paths and also the diffraction of the light through h slits.
1. How many bright interference fringes are within the central peak of the diffraction envelope
2. How many bright fringes are within either of the first side peaks of the diffraction envelope

Expert's answer

1. We can write for single slit

asinθ=mλ(1)a \sin θ = mλ (1)

where a is the slit of width, m=1

Using (1) we got

λ=asinθ(2)λ=a \sin θ (2)

We can write for bright interference fringes

dsinθ=kλ(3)d \sin θ = kλ (3)

Using (3) we got

k=dsinθλ(4)k=\frac{ d \sin θ }{λ} (4)

We put (2) in (4)

k=dsinθasinθ=da(5)k=\frac{ d \sin θ }{ a \sin θ }= \frac{ d }{ a } (5)

Using (5) we got

k=4

The number of bright interference fringes is equal to

0, ±1, ±2, ±3, ±4

Answer

9


2. The first side diffraction maximum can be located by setting m=1.5.

Using (1) we got

λ=asinθ1.5(6)λ=\frac{ a \sin θ }{1.5} (6)

We put (6) in (4)

k=1.5dsinθasinθ=1.5da(7)k=\frac{ 1.5 d \sin θ }{ a \sin θ }= \frac{ 1.5 d }{ a } (7)

Using (7) we got

k=7

The number of bright interference fringes is equal to

0, ±1, ±2, ±3, ±4, ±5, ±6, ±7

Answer

15


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