Question

The intensity of light with wavelength\({\bf{\lambda }}\)traveling through a diffraction grating with\({\bf{N}}\)slits at an angle\({\bf{\theta }}\) is given by \(I(\theta ) = {N^2}{\sin ^2}k/{k^2}\), where \({\bf{k = (\pi Ndsin\theta )/\lambda }}\)and\({\bf{d}}\) is the distance between adjacent slits. A helium-neon laser with wavelength \(\lambda = 632.8 \times {10^{ - 9}}\;{\rm{m}}\)is emitting a narrow band of light, given by\({\bf{ - 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{ < \theta < 1}}{{\bf{0}}^{{\bf{ - 6}}}}\), through a grating with 10,000 slits spaced \({\bf{1}}{{\bf{0}}^{{\bf{ - 4}}}}{\bf{\;m}}\)apart. Use the Midpoint Rule with \(n = 10\) to estimate the total light intensity \(\int_{{\bf{ - 1}}{{\bf{0}}^{{\bf{ - 6}}}}}^{{\bf{1}}{{\bf{0}}^{{\bf{ - 6}}}}} {\bf{I}} {\bf{(\theta )d\theta }}\) emerging from the grating.

Short answer

The value of\(\int_{ - {{10}^{ - 6}}}^{{{10}^{ - 6}}} I (\theta )d\theta \approx 59.41\).

Step-by-step solution

Given data

Given: \(I(\theta ) = \frac{{{N^2}{{\sin }^2}k}}{{{k^2}}} \cdots (1)\)

\(k = \frac{{(\pi Nd\sin \theta )}}{\lambda } \cdots (2)\)

\(N = 10000\,,\,d = {10^{ - 4}}\;{\rm{m}}\,,\,\lambda = 632.8 \times {10^{ - 9}}\;{\rm{m}}\)

Calculate value

Substitute value of \(N,d,\lambda \) to get the value of \(k\).

\(\begin{aligned}{c}k = \frac{{\left( {\pi \times 10000 \times {{10}^{ - 4}} \cdot \sin \theta } \right)}}{{632.8 \times {{10}^{ - 9}}}}\\k = \frac{{\pi \sin \theta }}{{632.8 \times {{10}^{ - 9}}}}\end{aligned}\)

Then from (1) we have

\(\begin{aligned}{l}I(\theta ) = {10^8} \cdot \frac{{{{\sin }^2}\left( {\frac{{\pi \sin \theta }}{{632.8 \times {{10}^{ - 9}}}}} \right)}}{{\frac{{{\pi ^2}{{\sin }^2}\theta }}{{{{\left( {632.8 \times {{10}^{ - 9}}} \right)}^2}}}}}\\I(\theta ) = \frac{{{{\left( {632.8 \times {{10}^{ - 9}}} \right)}^2} \times {{10}^8}{{\sin }^2}\left( {\frac{{\pi \sin \theta }}{{632.8 \times {{10}^{ - 9}}}}} \right)}}{{{\pi ^2}{{\sin }^2}\theta }} \ldots (3)\end{aligned}\)

Length of interval

It is given that\( - {10^{ - 6}} < \theta < {10^{ - 6}}\)

Divide interval in \(n = 10\)subintervals then

\(\begin{aligned}{c}\Delta \theta = \frac{{{{10}^{ - 6}} + {{10}^{ - 6}}}}{{10}}\\ = \frac{{2 \times {{10}^{ - 6}}}}{{10}}\\\Delta \theta = 2 \times {10^{ - 7}}\end{aligned}\)

Subintervals are\(\left( { - {{10}^{ - 6}}, - 0.8 \times {{10}^{ - 6}}} \right),\left( { - 0.8 \times {{10}^{ - 6}}, - 0.6 \times {{10}^{ - 6}}} \right),\left( { - 0.6 \times {{10}^{ - 6}}, - 0.4 \times {{10}^{ - 6}}} \right)\),

\(\begin{aligned}{l}\left( { - 0.4 \times {{10}^{ - 6}}, - 0.2 \times {{10}^{ - 6}}} \right),\left( { - 0.2 \times {{10}^{ - 6}},0} \right),\left( {0,0.2 \times {{10}^{ - 6}}} \right),\left( {0.2 \times {{10}^{ - 6}}),\left( {0.6 \times {{10}^{ - 6}},{{10}^{ - 6}}} \right)} \right.,\\\left( {0.4 \times {{10}^{ - 6}},0.6 \times {{10}^{ - 6}}} \right),\left( {0.6 \times {{10}^{ - 6}},0.8 \times {{10}^{ - 6}}} \right),\left( {0.8 \times {{10}^{ - 6}},\left( {0.8 \times {{10}^{ - 6}}} \right), - 0.8 \times {{10}^{ - 6}}, - 0.6 \times {{10}^{ - 6}}} \right)\end{aligned}\)

Use midpoint rule

Then midpoints are\( - 0.9 \times {10^{ - 6}}, - 0.7 \times {10^{ - 6}}, - 0.5 \times {10^{ - 6}}, - 0.3 \times {10^{ - 6}}, - 0.1 \times {10^{ - 6}},0.1 \times {10^{ - 6}}\), \(0.3 \times {10^{ - 6}},\;\;\;0.5 \times {10^{ - 6}},\;\;\;0.7 \times {10^{ - 6}},0.9 \times {10^{ - 6}}\)

By midpoints rule

\(\begin{aligned}{c}\int_{ - {{10}^{ - 6}}}^{{{10}^{ - 6}}} I (\theta )d\theta \approx {M_{10}}\\ = \Delta \theta \left( {I\left( { - 0.9 \times {{10}^{ - 6}}} \right) + I\left( { - 0.7 \times {{10}^{ - 6}}} \right) + \ldots \ldots ..I\left( {0.9 \times {{10}^{ - 6}}} \right)} \right)\\ = 0.2 \times {10^{ - 6}}\left( {I\left( { - 0.9 \times {{10}^{ - 6}}} \right) + I\left( { - 0.7 \times {{10}^{ - 6}}} \right) + \ldots \ldots .I\left( {0.9 \times {{10}^{ - 6}}} \right)} \right)\\ \approx 59.41\end{aligned}\)

\(\int_{ - {{10}^{ - 6}}}^{{{10}^{ - 6}}} I (\theta )d\theta \approx 59.41\)