Answer: The second interference is minimum at angle 5 degree approximately
Explanation:
Given that the
Wavelength (λ) = 633nm
Width d = 1.5 × 10^-5m
n = 2
The wavelength (λ) of the light passing through a single slit is related to the angle Ø by
dsinØ = n(λ)
Substitutes all the parameters into the above formula
1.5×10^-5 × SinØ = 2 × 633 × 10^-9
Make SinØ the subject of formula
SinØ = 1.266×10^-6/1.5×10^-5
SinØ = 0.0844
Ø = Sin^-1( 0.0844 )
Ø = 4.84 degree.
The second interference is minimum at angle 5 degree approximately
