Question

(a) Calculate the angle at which a 2.00-µm -wide slit produces its first minimum for 410-nm violet light. (b) Where is the first minimum for 700-nm red light?

Short answer

1. \({\rm{11}}{\rm{.8}}^\circ \)is the angle for the initial minimum for violet light.
2. \({\rm{20}}{\rm{.5}}^\circ \)is the angle for the initial minimum for the red light.

Step-by-step solution

Definition of wavelength

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

Given Data

Slit separation is\({\rm{D}} = 2.00\;\mu m\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{\;m}}}}{{1\;\mu m}}} \right)\; = {\rm{2}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{\;m}}\)

The wavelength of the light is,

\({\lambda _v} = 410\;nm\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 9}}}}{\rm{\;m}}}}{{1\;nm}}} \right) = 4.10 \times {\rm{1}}{{\rm{0}}^{{\rm{ - 7}}}}{\rm{\;m, }}{\lambda _r} = 700\;nm\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 9}}}}{\rm{\;m}}}}{{1\;nm}}} \right) = 7.00 \times {\rm{1}}{{\rm{0}}^{{\rm{ - 7}}}}{\rm{\;m}}\)

Find the angle at which a 2.00-µm -wide slit produces

To find the angle at which the violet light is at its first minimum, we utilize the equation\({\rm{D sin}}\left( {{{\rm{\theta }}_{\rm{v}}}} \right){\rm{ = m}}{{\rm{\lambda }}_{\rm{v}}}\)and solve it for\({{\rm{\theta }}_{\rm{v}}}\), we get

\(\begin{array}{c}{{\rm{\theta }}_{\rm{v}}} = {\rm{si}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {\frac{{{\rm{m}}{{\rm{\lambda }}_{\rm{v}}}}}{{\rm{D}}}} \right)\\ = {\rm{si}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {\frac{{{\rm{1}} \times 4.10 \times {\rm{1}}{{\rm{0}}^{{\rm{ - 7}}}}{\rm{\;m}}}}{{{\rm{2}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{\;m}}}}} \right)\\ = {\rm{11}}{\rm{.8}}^\circ \end{array}\)

Therefore, \({\rm{11}}{\rm{.8}}^\circ \) is the angle for the initial minimum for violet light.

Find the first minimum for 700-nm red light

We apply the equation to find the angle at which the red light is at its first minimum\({\rm{Dsin}}\left( {{{\rm{\theta }}_{\rm{r}}}} \right){\rm{ = m}}{{\rm{\lambda }}_{\rm{r}}}\)\({\rm{and solve it for}}\)\({{\rm{\theta }}_{\rm{r}}}\)

\(\begin{array}{c}{{\rm{\theta }}_{\rm{r}}}{\rm{ }} = {\rm{si}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {\frac{{{\rm{m}}{{\rm{\lambda }}_{\rm{r}}}}}{{\rm{D}}}} \right)\\ = {\rm{si}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {\frac{{{\rm{1}} \times {\rm{7}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 7}}}}{\rm{\;m}}}}{{{\rm{2}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{\;m}}}}} \right)\\ = {\rm{20}}{\rm{.5}}^\circ \end{array}\)

Therefore, \({\rm{20}}{\rm{.5}}^\circ \)is the angle for the initial minimum for the red light.