91Ó°ÊÓ

The wavelength of a waveform signal conveyed in space or down a wire is the distance between identical points (adjacent crests) in adjacent cycles.

To solve this problem, we solve it for theta using the equation\({\rm{D sin(\theta ) = m\lambda }}\), where\({\rm{m = 1 and solve it for \theta }}\).

\(\begin{array}{}{\rm{\theta = si}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {\frac{{{\rm{m\lambda }}}}{{\rm{D}}}} \right)\\{\rm{ = si}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {\frac{{{\rm{1 \times 550 \times 1}}{{\rm{0}}^{{\rm{ - 9}}}}{\rm{\;m}}}}{{{\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{\;m}}}}} \right)\\{\rm{ = 33}}{\rm{.4}}^\circ \end{array}\)

Therefore, \({\rm{33}}{\rm{.4}}^\circ \)is the required angle.

We calculate the equation\({\rm{Dsin(\theta ) = m\lambda }}\)for m=2 and check the angle for it to see if there is a second minimum.

\(\begin{array}{l}{\rm{sin(\theta ) = }}\frac{{{\rm{m\lambda }}}}{{\rm{D}}}\\{\rm{sin(\theta ) = }}\left( {\frac{{{\rm{2 \times 550 \times 1}}{{\rm{0}}^{{\rm{ - 9}}}}{\rm{\;m}}}}{{{\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{\;m}}}}} \right)\\{\rm{sin(\theta ) = 1}}{\rm{.1 }}\end{array}\)

The value of the sin function can not exceed maximum value 1, therefore this answer is invalid.

There can't be a second minimum because we got a value of\({\rm{sin(\theta )}}\)greater than 1.

Therefore, there will be no additional minimum.