Chapter 33, Problem 67P

To determine

The order of spectrum observe in transmitted light in range of $400\text{nm}$ to $700\text{nm}$ .

Explanation of Solution

Given:

The spacing of diffraction grating has $4800$ lines per centimeter.

The range of wavelength of transmitted light is $400\text{nm}$ to $700\text{nm}$ .

Formula used:

Write the formula for slit separation.

  $d=\frac{1}{N}$   ...... (1)

Here, $d$ is the distance between two slits and $N$ is the number of slits per centimeter.

Write the condition of maximum interface.

  $d\sin \theta =m\lambda$

Here, $d$ is the slit separation, $\theta$ is angle at which first spectrum expected, $m$ is the order of spectrum and $\lambda$ is the wavelength of line.

Rearrange the above equationfor $m$ .

  $m=\frac{d\sin \theta }{\lambda }$   ...... (2)

Rearrange the equation.

  $\sin \theta =\frac{m\lambda }{d}$

Rearrange the above equation for $\theta$ .

  $\theta ={\sin }^{-1}\left(\frac{m\lambda }{d}\right)$   ...... (3)

Calculation:

Substitute $4800$ per centimeter for $N$ in equation (1).

  $\begin{array}{c}d=\frac{1\text{cm}}{4800}\\ =0.00021\text{cm}\end{array}$

Substitute $1$ for $\sin \theta$ (maximum value of $\sin \theta$ ), $0.00021\text{cm}$ for $d$ and $400\text{nm}$ for $\lambda$ in equation (2).

  $\begin{array}{c}m=\frac{\left(0.00021\text{cm}\right)\left(1\right)}{400\text{nm}}\\ =\frac{\left(0.00021\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}{\left(400\text{nm}\right)\left(\frac{1\text{m}}{{10}^9\text{nm}}\right)}\\ \simeq 5\end{array}$

Substitute $1$ for $\sin \theta$ (maximum value of $\sin \theta$ ), $0.00021\text{cm}$ for $d$ and $700\text{nm}$ for $\lambda$ in equation (2).

  $\begin{array}{c}m=\frac{\left(0.00021\text{cm}\right)\left(1\right)}{700\text{nm}}\\ =\frac{\left(0.00021\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}{\left(700\text{nm}\right)\left(\frac{1\text{m}}{{10}^9\text{nm}}\right)}\\ \simeq 2\end{array}$

The order of spectrum in range $400\text{nm}$ to $700\text{nm}$ is $3\left(5-2\right)$ .

Substitute $3$ for $m$ , $400\text{nm}$ for $\lambda$ and $0.00021\text{cm}$ for $d$ in equation (3).

  $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(3\right)\left(400\text{nm}\right)}{\left(0.00021\text{cm}\right)}\right)\\ ={\sin }^{-1}\left(\frac{\left(3\right)\left(400\text{nm}\right)\left(\frac{1\text{m}}{{10}^9\text{nm}}\right)}{\left(0.00021\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\right)\\ ={\sin }^{-1}\left(0.576\right)\\ =35.17°\end{array}$

Substitute $4$ for $m$ , $400\text{nm}$ for $\lambda$ and $0.00021\text{cm}$ for $d$ in equation (3).

  $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(4\right)\left(400\text{nm}\right)}{\left(0.00021\text{cm}\right)}\right)\\ ={\sin }^{-1}\left(\frac{\left(4\right)\left(400\text{nm}\right)\left(\frac{1\text{m}}{{10}^9\text{nm}}\right)}{\left(0.00021\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\right)\\ ={\sin }^{-1}\left(0.768\right)\\ =50.174°\end{array}$

Substitute $5$ for $m$ , $400\text{nm}$ for $\lambda$ and $0.00021\text{cm}$ for $d$ in equation 3.

  $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(5\right)\left(400\text{nm}\right)}{\left(0.00021\text{cm}\right)}\right)\\ ={\sin }^{-1}\left(\frac{\left(5\right)\left(400\text{nm}\right)\left(\frac{1\text{m}}{{10}^9\text{nm}}\right)}{\left(0.00021\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\right)\\ ={\sin }^{-1}\left(0.960\right)\\ =73.739°\end{array}$

Angle for $\lambda =700\text{nm}$ are out of range. So these order overlap with each other and the overlapping regions will be for $\lambda >400\text{nm}$ .

Conclusion:

Thus, the order of spectrum range from $400\text{nm}$ to $700\text{nm}$ is $5$ and there will be overlap of long wavelengths of second order spectrum with the short wavelengths of third order spectrum.