Answer:
Angles of the first two diffraction orders are [tex]3.152^0 , 6.324^0[/tex]
Explanation:
Number of slit N=1000
wavelength [tex]\lambda=550 nm[/tex]
width of diffraction grating = 1 cm
Distance between the any two adjacent slit is
[tex]d=\frac0{1}{N} \\d=\frac{1\times 10^{-2}}{1000} =10^{-5} m[/tex]
angle of nth fringe is given by
[tex]sin\theta_{n}=\frac{n\lambda}{d}[/tex]
for first order n=1, which is given by
[tex]sin\theta_{1}=\frac{1\lambda}{d} \\\theta_{1}=sin^{-1}(\frac{(1\times 550 \times 10^{-9})}{ 10^{-5}} )\\\theta_{1}=3.152^0[/tex]
Similarly for 2nd order
[tex]sin\theta_{2}=\frac{2\times \lambda}{d} \\\theta_{2}=sin^{-1}(\frac{(2\times 550 \times 10^{-9})}{ 10^{-5}} )\\\theta_{2}=6.324^0[/tex]
By using the Huygens formula, the angles of the first two orders are 6.3 degrees and 13 degrees approximately.
Given that a 1.0cm wide diffraction grating has 1000 slits.
The slit spacing = (1.0 x [tex]10^{-2}[/tex]) / 1000
The slit spacing = 1.0 x [tex]1.6^{-5}[/tex]
Also, it is given that It is illuminated by light of wavelength 550 nm
Sin∅ = nλ/d
When n = 1
Sin∅ = 550 x [tex]10^{-9}[/tex]/1.0 x [tex]10^{-5}[/tex]
Sin∅ = 0.055
∅ = [tex]Sin^{-1}[/tex] ( 0.055)
∅ = 3.152
The angle of the first order = 2∅ = 2 x 3.152 = 6.305
2∅ = 6.3 degrees approximately
When n = 2
Sin∅ = nλ/d
Sin∅ = 2 x 550 x [tex]10^{-9}[/tex]/1.0 x [tex]10^{-5}[/tex]
Sin∅ = 0.11
∅ = [tex]Sin^{-1}[/tex] ( 0.11)
∅ = 6.3153
The angle of the second order = 2∅
2∅ = 6.3153 x 2
2∅ = 13 degrees approximately
Therefore, the angles of the first two orders are 6.3 degrees and 13 degrees approximately.
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