As per the given question,
Wavelength of the wave (λ)=550nm(\lambda)=550nm(λ)=550nm
θ=23∘\theta=23^\circθ=23∘
sinθ=nλd\sin \theta =n\frac{\lambda}{d}sinθ=ndλ
from bright fringe n = 1
Now, substituting the values
sin23∘=1×550×10−9md\sin 23^\circ =1\times \frac{550\times 10^{-9}m}{d}sin23∘=1×d550×10−9m
⇒d=5.5×10−90.39\Rightarrow d = \frac{5.5\times 10^{-9}}{0.39}⇒d=0.395.5×10−9
⇒d=14.1×10−9m\Rightarrow d = 14.1\times 10^{-9}m⇒d=14.1×10−9m
d=0.0141μmd=0.0141\mu md=0.0141μm
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