Answer:
(a) 3.9 mm
(b) 5.85 mm
Explanation:
Given that:
d = 0.500 mm, we need to convert this to meters so, d = 5.0 × [tex]10^{-4}[/tex] m
lambda = 650 nm, this has to be converted to meters too so,
lambda = 6.5 × [tex]10^{-7}[/tex] m and L = 3.00 m
(the symbol for lambda is not available so I'm just using the worded form)
(a) the general formula for finding the interference pattern is
dsin(theta) = nlambda
where sin(theta) = [tex]\frac{y}{L}[/tex] since the angle is not provided and n = 1 here since it refers to first bright fringe
(5.0 × [tex]10^{-4}[/tex])([tex]\frac{y}{3}[/tex]) = (1)(6.5 × [tex]10^{-7}[/tex])
[tex]\\ y[/tex] = [(3)(6.5 × [tex]10^{-7}[/tex])] ÷ (5.0 × [tex]10^{-4}[/tex])
[tex]\\ y[/tex] = 0.0039 m = 3.9 mm
(b) for the dark fringe, the formula will now be
dsin(theta) = (n-0.5)lambda
(5.0 × [tex]10^{-4}[/tex])([tex]\frac{y}{3}[/tex]) = (2-0.5)(6.5 × [tex]10^{-7}[/tex])
[tex]\\ y[/tex] = [(3)(1.5)(6.5 × [tex]10^{-7}[/tex])] ÷ (5.0 × [tex]10^{-4}[/tex])
[tex]\\ y[/tex] = 0.00585 m = 5.85 mm
Hope that answers the question. Have a great day!
