(a) What is the width of a single slit that produces its first minimum at \({\rm{60}}{\rm{.0^\circ }}\) for 600-nm light? (b) Find the wavelength of light that has its first minimum at \({\rm{62}}{\rm{.}}{{\rm{0}}^{\rm{^\circ }}}\).

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

Angle for the first minimum is\({\rm{60}}{\rm{.0}}^\circ \)

Wavelength of the light is,

\(\lambda = 600\;nm\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 9}}}}{\rm{\;m}}}}{{1\;nm}}} \right) = 6.00 \times {\rm{1}}{{\rm{0}}^{{\rm{ - 7}}}}{\rm{\;m}}\)

To find the width of the slit, solve\({\rm{Dsin(\theta ) = m\lambda }}\)for D, where\({\rm{m = 1}}\),

\(\begin{array}{c}{\rm{D = }}\frac{{{\rm{m\lambda }}}}{{{\rm{sin(\theta )}}}}\\{\rm{ = }}\frac{{{\rm{1 \times 6}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ - 7}}}}{\rm{\;m}}}}{{{\rm{sin}}\left( {{\rm{60}}{\rm{.0}}^\circ } \right)}}\\{\rm{ = 6}}{\rm{.93 \times 1}}{{\rm{0}}^{{\rm{ - 7}}}}{\rm{\;m}}\end{array}\)

Therefore, the slit's width is \({\rm{6}}{\rm{.93 \times 1}}{{\rm{0}}^{{\rm{ - 7}}}}{\rm{\;m}}\).

Let us calculate the equation\({\rm{Dsin(\theta ) = m\lambda }}\)for lambda where m=1 to get the wavelength for the initial minimum.

\(\begin{array}{c}{\rm{\lambda = }}\frac{{{\rm{Dsin(\theta )}}}}{{\rm{m}}}\\{\rm{ = }}\frac{{{\rm{6}}{\rm{.93 \times 1}}{{\rm{0}}^{{\rm{ - 7}}}}{\rm{\;m \times sin}}\left( {{\rm{62}}{\rm{.0}}^\circ } \right)}}{{\rm{1}}}\\{\rm{ = 6}}{\rm{.12 \times 1}}{{\rm{0}}^{{\rm{ - 7}}}}{\rm{\;m}}\end{array}\)

Hence, the wavelength of light that has its initial minimum at \({\rm{62}}{\rm{.0}}^\circ \)is \({\rm{6}}{\rm{.12 \times 1}}{{\rm{0}}^{{\rm{ - 7}}}}{\rm{\;m}}\).