1. We can write for diffraction minimum
"a \\sin \u03b8 = m\u03bb (1)"
where a is the slit of width
Using (1) we got
"a=\\frac{ m\u03bb }{\\sin \u03b8 } (2)"
In our case, θ=15°, m=1, λ=650×10-9 m
Using (2) we got
a=2.5×10-6 m
Answer
2.5×10-6 m
2. The first side diffraction maximum can be located by setting m=1.5.
In this case, we can rewrite (1) as
"a \\sin \u03b8 = 1.5\u03bb (2)"
In our case, a=2.5×10-6 m, θ=15°
Using (2) we got
λ=430×10-9 m
Answer
430×10-9 m
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