Answer:
width of slit =1.23× 10⁻⁶ m
Explanation:
we know the condition of diffraction minima,
d sin θ = n λ
λ = wavelength θ = angle between the central maxima and 1st minima
d = slit width
for first minima n = 1
now,
d =[tex]\dfrac{n \lambda}{sin \theta}[/tex]
[tex]\theta = \dfrac{45^0}{2} = 22.5^0[/tex]
d =[tex]\dfrac{1\times 460 \times 10^{-9}}{sin 22.5^0}[/tex]
d = 1228 × 10⁻⁹ m = 1.228× 10⁻⁶ m
d = 1.23× 10⁻⁶ m
width of slit =1.23× 10⁻⁶ m
