To solve this problem, the concepts related to single slit diffraction are necessary. The expression for the diffraction minimum in the single slit diffraction is
[tex]dsin\theta = m \lambda[/tex]
Where,
d= Slit width
m = Order of the diffraction
\lambda = Wavelength
The wavelength can be calculated through the expression
[tex]\lambda= \frac{c}{f}[/tex]
Where,
c = speed of light
f = Frequency
Note: The angular width of the electromagnetic wave that is width of the central maximum is equal to the twice of the angle of diffraction for the first minimum.
[tex]\theta'=2\theta[/tex]
Therefore, applying to our values we have,
[tex]\lambda= \frac{c}{f}[/tex]
[tex]\lambda = \frac{3.0*10^8}{833*10^6}[/tex]
[tex]\lambda =0.3601m[/tex]
To calculate the angle we can use the expression for diffraction and re-arrange for \theta, therefore
[tex]dsin\theta = m \lambda[/tex]
[tex]\theta = sin^{-1}(\frac{(1)0.3601}{16})[/tex]
[tex]\theta = 1.2896[/tex]
The angular widht would be
[tex]\theta'=2\theta[/tex]
[tex]\theta'=2(1.2896)[/tex]
[tex]\theta' = 2.58\°[/tex]
Therefore the angular width of the electromagnetic wave is [tex]2.58\°[/tex]
