Answer:
The first diffraction maximum fringe will be at approximately 2.7 meters from the central maximum.
Explanation:
We can describe single slit diffraction phenomenon with the equation:
[tex] a\sin\theta=m\lambda[/tex] (1)
with θ the angular position of the minimum of order m respect the central maximum, a the slit width and λ the wavelength of the incident light. Because the distances between the first minima and the central maximum ([tex] y_{m} [/tex]) are small compared to the distance between the screen and the slit (x), we can approximate [tex] \sin\theta\approx\tan\theta=\frac{y}{x} [/tex], using this on (1):
[tex]a\frac{y_{m}}{x}=m\lambda [/tex]
solving for y
[tex]y_{m}= \frac{mx\lambda}{a} [/tex]
Note that [tex]y_{m} [/tex]is the distance between a minimum and the central maximum but we need the position of a maximum not a minimum, here we can use the fact that a maximum is approximately between two minima, so the first diffraction maximum fringe is between the minima of order 1 and 2, so we should find [tex] y_{1} [/tex], [tex] y_{2} [/tex] add them and divide by two:
[tex] y_{1}= \frac{(1)(10.0m)(550\times10^{-9}\,m)}{3.00\times10^{-6}\,m}[/tex]
[tex] y_{1}= 1.8 m[/tex]
[tex] y_{2}= \frac{(2)(10.0m)(550\times10^{-9}\,m)}{3.00\times10^{-6}\,m}[/tex]
[tex] y_{1}= 3.6 m[/tex]
[tex] maximum = \frac{1.8+3.6}{2}=2.7m[/tex]
