Chapter 38, Problem 38.31P

A diffraction grating has 4 200 rulings/cm. On a 2.00 m from the grating. it is found that for a particular  order m, the maxima corresponding lo two closely spaced wavelengths of sodium (589.6 nm and 589.6 nm) are separated by 1.54 mm. Determine the value of m.

To determine

The value of $m$.

Answer to Problem 38.31P

The value of $m$ is $2$.

Explanation of Solution

Given info: The number a grating is $4200\text{rulings}/\text{cm}$. The distance between screen and grating is $2.00\text{m}$. The maxima corresponding to $589.0\text{nm}$ and $589.6\text{nm}$ are separated by $1.54\text{mm}$.

The first wavelength of sodium is $589.0\text{nm}$ and second wavelength of sodium is $589.6\text{nm}$.

For the first order the value of $m$ is $1$.

The formula to calculate separation between gratings is,

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

Here,

$N$ is number of rulings on diffraction grating.

Substitute $4200\text{rulings}/\text{cm}$ for $N$ in the above expression.

$\begin{array}{c}d=\frac{1}{4200\text{rulings}/\text{cm}}\\ =\frac{1}{4200\text{rulings}/\text{cm}\times \left(\frac{1\text{cm}}{{10}^{-2}\text{m}}\right)}\\ =2380\times {10}^{-9}\text{m}\times \left(\frac{{10}^9\text{nm}}{1\text{m}}\right)\\ =2380\text{nm}\end{array}$

It remains same for all the order.

The formula for diffraction grating is,

$d\sin {\theta }_1=m{\lambda }_1$

Here,

${\theta }_1$ is angle corresponding to first wavelength.

${\lambda }_1$ is first wavelength.

Rearrange the above expression for ${\theta }_1$,

${\theta }_1={\sin }^{-1}\frac{m{\lambda }_1}{d}$

Substitute $589.0\text{nm}$ for ${\lambda }_1$, $2380\text{nm}$ for $d$ and $1$ for $m$ in the above expression.

$\begin{array}{c}{\theta }_1={\sin }^{-1}\frac{\left(1\right)\left(589.0\text{nm}\right)}{2380\text{nm}}\\ =14.32°\end{array}$

Write the expression for tan of the above angle,

$\tan {\theta }_1$

Substitute $14.32°$ for ${\theta }_1$ in the above expression.

$\begin{array}{c}{\theta }_2=\tan \left(14.32°\right)\\ =0.255268582°\end{array}$

The formula for diffraction grating is for second wavelength is,

$d\sin {\theta }_3=m{\lambda }_2$

Here,

${\theta }_3$ is angle corresponding to second wavelength.

${\lambda }_2$ is second  wavelength.

Rearrange the above expression for ${\theta }_3$,

${\theta }_3={\sin }^{-1}\frac{m{\lambda }_2}{d}$

Substitute $589.6\text{nm}$ for ${\lambda }_3$, $2380\text{nm}$ for $d$ and $1$ for $m$ in the above expression.

$\begin{array}{c}{\theta }_3={\sin }^{-1}\frac{\left(1\right)\left(589.6\text{nm}\right)}{2380\text{nm}}\\ =14.34°\end{array}$

Write the expression for tan of the above angle,

${\theta }_4=\tan {\theta }_2$

Substitute $14.34°$ for ${\theta }_3$ in the above expression.

$\begin{array}{c}{\theta }_4=\tan \left(14.34°\right)\\ =0.255640427°\end{array}$

The difference between the tan of the two angle is,

${\theta }_5={\theta }_4-{\theta }_3$

Substitute $0.255640427°$ for ${\theta }_4$ and $0.255268582°$ for ${\theta }_3$ in the above expression.

$\begin{array}{c}{\theta }_5=\left(0.255640427°\right)-\left(0.255268582°\right)\\ =3.71845\times {10}^{-4}\end{array}$

Write the formula to calculate position of central $m^{\text{th}}$ maxima from central maxima is,

$y=L\tan {\theta }_5$

Here,

$L$ is distance between grating and screen.

Substitute $2.00\text{m}$ for $L$ and $3.71845\times {10}^{-4}$ for ${\theta }_5$ in the above expression.

$\begin{array}{c}y=\left(2.00\text{m}\right)\left(3.71845\times {10}^{-4}\right)\\ =7.436\times {10}^{-4}\text{m}\times \left(\frac{{10}^3\text{mm}}{1\text{m}}\right)\\ =0.7436\text{mm}\end{array}$

For the second order the value of $m$ is $2$.

The formula for diffraction grating is,

$d\sin {\theta }_6=m{\lambda }_1$

Here,

${\theta }_6$ is angle corresponding to first wavelength and second order.

${\lambda }_1$ is first wavelength.

Rearrange the above expression for ${\theta }_1$,

${\theta }_6={\sin }^{-1}\frac{m{\lambda }_1}{d}$

Substitute $589.0\text{nm}$ for ${\lambda }_1$, $2380\text{nm}$ for $d$ and $2$ for $m$ in the above expression.

$\begin{array}{c}{\theta }_6={\sin }^{-1}\frac{\left(2\right)\left(589.0\text{nm}\right)}{2380\text{nm}}\\ ={\sin }^{-1}\left(0.49495°\right)\\ =29.66°\end{array}$

Write the expression for tan of the above angle,

${\theta }_7=\tan {\theta }_6$

Substitute $29.66°$ for ${\theta }_6$ in the above expression.

$\begin{array}{c}{\theta }_7=\tan \left(29.66°\right)\\ =0.569465032°\end{array}$

The formula for diffraction grating is for second wavelength is,

$d\sin {\theta }_8=m{\lambda }_2$

Here,

${\theta }_8$ is angle corresponding to second wavelength and the order is $2$.

${\lambda }_2$ is second  wavelength.

Rearrange the above expression for ${\theta }_8$,

${\theta }_8={\sin }^{-1}\frac{m{\lambda }_2}{d}$

Substitute $589.6\text{nm}$ for ${\lambda }_2$, $2380\text{nm}$ for $d$ and $2$ for $m$ in the above expression.

$\begin{array}{c}{\theta }_8={\sin }^{-1}\frac{\left(2\right)\left(589.6\text{nm}\right)}{2380\text{nm}}\\ ={\sin }^{-1}\left(0.49546°\right)\\ =29.7°\end{array}$

Write the expression for tan of the above angle,

${\theta }_9=\tan {\theta }_8$

Substitute $29.7°$ for ${\theta }_8$ in the above expression.

$\begin{array}{c}{\theta }_9=\tan \left(29.7°\right)\\ =0.570389929°\end{array}$

The difference between the tan of the two angle is,

${\theta }_{10}={\theta }_9-{\theta }_7$

Substitute $0.570389929°$ for ${\theta }_9$ and $0.569465032°$ for ${\theta }_7$ in the above expression.

$\begin{array}{c}{\theta }_{10}=\left(0.570389929°\right)-\left(0.569465032°\right)\\ =9.24897\times {10}^{-4}\end{array}$

Write the formula to calculate position of central $m^{\text{th}}$ maxima from central maxima is,

$y=L{\theta }_{10}$

Here,

$L$ is distance between grating and screen.

Substitute $2.00\text{m}$ for $L$ and $9.24897\times {10}^{-4}$ for ${\theta }_{10}$ in the above expression.

$\begin{array}{c}y=\left(2.00\text{m}\right)\left(9.24897\times {10}^{-4}\right)\\ =1.8497\times {10}^{-3}\text{m}\times \left(\frac{{10}^3\text{mm}}{1\text{m}}\right)\\ =1.8497\text{mm}\end{array}$

For the third order the value of $m$ is $3$.

The formula for diffraction grating is,

$d\sin {\theta }_{11}=m{\lambda }_1$

Here,

${\theta }_{11}$ is angle corresponding to first wavelength and second order.

${\lambda }_1$ is first wavelength.

Rearrange the above expression for ${\theta }_{11}$,

${\theta }_{11}={\sin }^{-1}\frac{m{\lambda }_1}{d}$

Substitute $589.0\text{nm}$ for ${\lambda }_1$, $2380\text{nm}$ for $d$ and $3$ for $m$ in the above expression.

$\begin{array}{c}{\theta }_{11}={\sin }^{-1}\frac{\left(3\right)\left(589.0\text{nm}\right)}{2380\text{nm}}\\ ={\sin }^{-1}\left(0.7424°\right)\\ =47.936°\end{array}$

Write the expression for tan of the above angle,

${\theta }_{12}=\tan {\theta }_{11}$

Substitute $47.936°$ for ${\theta }_{11}$ in the above expression.

$\begin{array}{c}{\theta }_{12}=\tan \left(47.936°\right)\\ =1.108120806°\end{array}$

The formula for diffraction grating is for second wavelength is,

$d\sin {\theta }_{13}=m{\lambda }_2$

Here,

${\theta }_{13}$ is angle corresponding to second wavelength and the order is $3$

${\lambda }_2$ is second  wavelength.

Rearrange the above expression for ${\theta }_8$,

${\theta }_{13}={\sin }^{-1}\frac{m{\lambda }_2}{d}$

Substitute $589.6\text{nm}$ for ${\lambda }_2$, $2380\text{nm}$ for $d$ and $3$ for $m$ in the above expression.

$\begin{array}{c}{\theta }_{13}={\sin }^{-1}\frac{\left(3\right)\left(589.6\text{nm}\right)}{2380\text{nm}}\\ ={\sin }^{-1}\left(0.7431°\right)\\ =47.99°\end{array}$

Write the expression for tan of the above angle,

${\theta }_{14}=\tan {\theta }_{13}$

Substitute $29.7°$ for ${\theta }_8$ in the above expression.

$\begin{array}{c}{\theta }_{14}=\tan \left(47.99°\right)\\ =1.110222778°\end{array}$

The difference between the tan of the two angle is,

${\theta }_{15}={\theta }_{14}-{\theta }_{12}$

Substitute $1.110222778°$ for ${\theta }_{14}$ and $1.108120806°$ for ${\theta }_{12}$ in the above expression.

$\begin{array}{c}{\theta }_{15}=\left(1.110222778°\right)-\left(1.108120806°\right)\\ =2.101972\times {10}^{-3}\end{array}$

Write the formula to calculate position of central $m^{\text{th}}$ maxima from central maxima is,

$y=L{\theta }_{15}$

Here,

$L$ is distance between grating and screen.

Substitute $2.00\text{m}$ for $L$ and $2.101972\times {10}^{-3}$ for ${\theta }_{15}$ in the above expression.

$\begin{array}{c}y=\left(2.00\text{m}\right)\left(2.101972\times {10}^{-3}\right)\\ =4.2039\times {10}^{-3}\text{m}\times \left(\frac{{10}^3\text{mm}}{1\text{m}}\right)\\ =4.203\text{mm}\end{array}$

From the above calculation it is clear that only for second order the maxima of two closely spaced wavelength of sodium $589.0\text{nm}$ and $589.6\text{nm}$ are separated by $1.54\text{mm}$.

Conclusion:

Therefore, the value of $m$ is $2$.