Answer:
The distance that the diffraction spot moved on the screen is 0.314 cm
Explanation:
The condition for maxima is equal:
[tex]sin\theta _{max} =\frac{\lambda }{D*1.33}\\\theta _{max} =sin^{-1} \frac{\lambda }{D*1.33}[/tex]
Where
λ = 650 nm = 650x10⁻⁹m
D = distances = 1.87 μm = 1.87x10⁻⁶m and 1.94x10⁻⁶m
Replacing:
[tex]\theta _{1.87} =sin^{-1} \frac{650x10^{-9} }{1.87x10^{-6}*1.33 } =15.15[/tex]
[tex]\theta _{1.94} =sin^{-1} \frac{650x10^{-9} }{1.94x10^{-6}*1.33 } =14.59[/tex]
The distance that the diffraction spot moved on the screen is equal to:
[tex]x=d*(tan\theta _{1.87} -tan\theta _{1.94} )=30*(tan15.15-tan14.59)=0.314cm[/tex]
