Question

With light from a gaseous discharge tube incident normally on a grating with slit separation $\mathbf{1}\mathbf{.}\mathbf{73}\mathit{\mu }\mathit{m}$$\mathbf{37}\mathbf{.}\mathbf{3}\mathbf{°}\mathbf{,}$$\mathbf{-}\mathbf{37}\mathbf{.}\mathbf{1}\mathbf{°}\mathbf{,}$$\mathbf{65}\mathbf{.}\mathbf{2}\mathbf{°}$, sharp maxima of green light are experimentally found at angles localid="1664274691722" $\mathit{\theta }\mathbf{=}\mathbf{\pm }\mathbf{17}\mathbf{.}\mathbf{6}\mathbf{°}\mathbf{,}$and $\mathbf{-}\mathbf{65}\mathbf{.}\mathbf{0}\mathbf{°}$. Compute the wavelength of the green light that best fits these data.

Short answer

The best fit wavelength is, 523 nm.

Step-by-step solution

Step 1: Identification of the given data

The slit separation,$d=1.73\mu m$

The angles,$\theta =\pm 17.6°,37.3°,-37.1°,65.2°,-65°$

Determine the formulas to calculate the wavelength of the green light.

The condition for the maxima in diffraction to calculate the wavelength of the green light is given as follows.

$$\begin{gathered} dsin\theta =m\lambda \\ \lambda =\frac{dsin\theta }{m} \end{gathered}$$ .........(1)

Here, is the order of the diffraction and $\mathit{\lambda }$is the wavelength.

Calculate the wavelength of the green light.

Calculate the wavelength for $\theta =-17.6°$and $m=1$.

Substitute $1.73\mathrm{\mu m}$for $d$, $-17.6°$ for $\theta$, and 1 for$m$into equation (i).
$$\begin{gathered} \lambda =\frac{\left|1.73\times {10}^{-6} \text{m}\times \sin \left(-17.6\right)\right|}{1} \\ \lambda =0.523\times {10}^{-6}\text{m} \\ \lambda =\text{523}\text{nm} \end{gathered}$$

Calculate the wavelength for $17.6°$and $m=1$

Substitute$1.73\mu m$for$d$, $17.6°$for $\theta$, and 1 for m into equation (i).

$$\begin{gathered} \lambda =\frac{\left|1.73\times {10}^{-6}\text{m}\times \sin \left(17.6\right)\right|}{1} \\ \lambda =0.523\times {10}^{-6}\text{m} \\ \lambda =\text{523}\text{nm} \end{gathered}$$

Calculate the wavelength $\theta =37.3°$for and $m=2$.

Substitute $1.73\mu m$for $d$, $37.3°$for $\theta$, and 1 for $m$into equation (i).

$$\begin{gathered} \lambda =\frac{\left|1.73\times {10}^{-6}\text{m}\times \sin \left(37.3\right)\right|}{2} \\ \lambda =0.524\times {10}^{-6}\text{m} \\ \lambda =\text{524}\text{nm} \end{gathered}$$

Calculate the wavelength for$\theta =-37.1°$and $m=2$.

Substitute $1.73\mu m$for$d$, $-37.1°$for$\theta$, and 2 for $m$into equation (i).

$$\begin{gathered} \lambda =\frac{\left|1.73\times {10}^{-6}\text{m}\times \sin \left(-37.1\right)\right|}{2} \\ \lambda =0.522\times {10}^{-6}\text{m} \\ \lambda =\text{522}\text{nm} \end{gathered}$$

Calculate the wavelength for$\theta =65.2°$and $m=3$.

Substitute $1.73\mu m$for $d$, $65.2°$for $\theta$, and 3 for $m$into equation

(i).

$$\begin{gathered} \lambda =\frac{\left|1.73\times {10}^{-6}\text{m}\times \sin \left(65.2\right)\right|}{3} \\ \lambda =0.523\times {10}^{-6}\text{m} \\ \lambda =\text{523}\text{nm} \end{gathered}$$

Calculate the wavelength for $\theta =-65.°$and $m=3$.

Substitute $1.73\mu m$for $d$, $-65°$for $\theta$, and 3 for $m$into equation (i).

$$\begin{gathered} \lambda =\frac{\left|1.73\times {10}^{-6}\text{m}\times \sin \left(-65\right)\right|}{3} \\ \lambda =0.523\times {10}^{-6}\text{m} \\ \lambda =\text{523}\text{nm} \end{gathered}$$

The average of the wavelengths to calculate the best wavelength,

$$\begin{gathered} \lambda =\frac{523\text{nm}+524\text{nm}+522\text{nm}+523\text{nm}+523\text{nm}}{5} \\ \lambda =523\text{nm} \end{gathered}$$

Hence the best fit wavelength is, 523 nm.