Answer
$\alpha = 2.5mm$
Work Step by Step
It is known that:
The distance between the first and fifth minima of a single slit diffraction pattern is 0.35 mm with the screen 40 cm away from the slit, when light of wavelength 550 nm is used.
From the triangle, we know:
$\Delta y=D \sin \theta$,
where $\Delta y$ is the distance between minima and D is the distance to the screen.
Solved for $\sin \theta$, he have:
$(\Delta y)/D= \sin \theta$,
From equation (36-3), we know:
$\alpha \Delta \sin \theta = \Delta m \lambda $ where $m=1,2,3,....$ (minima)
Solving for alpha:
$\alpha =( \Delta m \lambda )/(\sin \theta)$
Substituting for $\sin \theta$, we have:
$\alpha =( \Delta m \lambda )/(\Delta y/D)$
$\alpha =(5-1) 550*10^{-9m} /(.35*10^-3m/0.40m)$
$\alpha = 2.5*10^-3m$
$\alpha = 2.5mm$
