Question

A) What is the wavelength of light that is deviated in the first order through an angle of 14.1 degrees by a transmission grating having 5000 slits/cm? Assume normal incidence.

B)What is the second-order deviation of this wavelength? Assume normal incidence.

Answer 1

Concepts and reason

The concept used to solve this problem is transmission grating.

Initially, the separation between the slits can be calculated from the number of slits of the grating. Then, the wavelength of the first order deviation can be calculated by using the expression of Bragg’s law. Finally, the deviation angle of the second order deviation can be calculated using the expression of Bragg’s law.

Fundamentals

The expression for the separation between slits is given below:

$d = \frac{1}{N}$

Here, $d$is the separation distance, $N$is the number of slits of the grating.

The expression for the wavelength from Bragg’s law is given below:

$d\sin \theta = n\lambda$

Here, $\theta$is the angle, $n$is the order of deviation, $\lambda$is the wavelength.

(A)

The expression for the separation distance is given below:

$d = \frac{1}{N}$

Substitute $5000\;{\rm{slits/cm}}$for$N$.

$\begin{array}{c}\\d = \frac{1}{{\left( {5000\;{\rm{slits/cm}}} \right)\left( {\frac{{1\;{\rm{cm}}}}{{1 \times {{10}^{ - 2}}\;{\rm{m}}}}} \right)}}\\\\ = 2 \times {10^{ - 6}}{\rm{m}}\\\end{array}$

The expression for the wavelength from Bragg’s law is given below:

$\begin{array}{c}\\d\sin \theta = n\lambda \\\\\lambda = \frac{{d\sin \theta }}{n}\\\end{array}$

Substitute $2 \times {10^{ - 6}}{\rm{m}}$for$d$, $14.1^\circ$for$\theta$, $1$for$n$

$\begin{array}{c}\\\lambda {\rm{ = }}\frac{{\left( {2 \times {{10}^{ - 6}}\;{\rm{m}}} \right)\sin \left( {14.1^\circ } \right)}}{1}\\\\ = 4.87 \times {10^{ - 7}}{\rm{m}}\\\end{array}$

(B)

The expression for the deviation angle from Bragg’s law is given below:

$\begin{array}{c}\\d\sin \theta = n\lambda \\\\\theta = {\sin ^{ - 1}}\left( {\frac{{n\lambda }}{d}} \right)\\\end{array}$

Substitute $2 \times {10^{ - 6}}{\rm{m}}$for$d$,$4.87 \times {10^{ - 7}}{\rm{m}}$for $\lambda$ and$2$for$n$.

$\begin{array}{c}\\\theta = {\sin ^{ - 1}}\left( {\frac{{\left( 2 \right)\left( {4.87 \times {{10}^{ - 7}}{\rm{m}}} \right)}}{{\left( {2 \times {{10}^{ - 6}}\;{\rm{m}}} \right)}}} \right)\\\\ = 29.14^\circ \\\end{array}$

Ans: Part A

The wavelength of the first order deviation is ${\bf{4}}{\bf{.87 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}\;{\bf{m}}$.

Part B

The deviation angle of the second order deviation is ${\bf{29}}{\bf{.14^\circ }}$.