Answer:
a)[tex]y_{first}=5.3mm[/tex]
b)[tex]y_{second}=10.6-5.3 =5.3 mm[/tex]
Explanation:
a)
The width of the central bright in this diffraction pattern is given by:
[tex]y=\frac{m\lambda D}{a}[/tex] when m is a natural number.
here:
So we have:
[tex]y_{first}=\frac{633*10^{9}*3.35}{0.0004}[/tex]
[tex]y_{first}=5.3mm[/tex]
b)
Now, if we do m=2 we can find the distance to the second minima.
[tex]y_{2}=\frac{2*633*10^{9}*3.35}{0.0004}[/tex]
[tex]y_{2}=10.6 mm[/tex]
Now we need to subtract these distance, to get the width of the first bright fringe :
[tex]y_{second}=10.6-5.3 =5.3 mm[/tex]
I hope it heps you!