Answer
given,
width of slit, d = 0.08 mm
d = 8 x 10⁻⁵ m
light of two wavelength
I₁= 446 nm
I₂ = 662 nm
a) angles at which the third dark fringe
[tex]sin C = \dfrac{m\lambda}{d}[/tex]
m = 3 , I₁= 446 nm
[tex]sin C = \dfrac{3\times 446 \times 10^{-9}}{8\times 10^{-5}}[/tex]
C = 0.958°
m = 3 , I₁= 662 nm
[tex]sin C = \dfrac{3\times 662 \times 10^{-9}}{8\times 10^{-5}}[/tex]
C = 1.423°
b) angles at which the third dark fringe
[tex]sin C = \dfrac{m\lambda}{d}[/tex]
m = 1 , I₁= 446 nm
[tex]sin C = \dfrac{1\times 446 \times 10^{-9}}{8\times 10^{-5}}[/tex]
C = 0.319°
m = 1 , I₁= 662 nm
[tex]sin C = \dfrac{1\times 662 \times 10^{-9}}{8\times 10^{-5}}[/tex]
C = 0.474°