Chapter 23, Problem 26E

(a)

To determine

Angular position for $m=1$

Answer to Problem 26E

The angular position for $m=1$ is $13.6°$

Explanation of Solution

Write the expression for double slit interference.

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

Here, $\lambda$ is the wavelength of light, $m$ is the order of interference, $d$ is the distance between the slits and $\theta$ is the angle of the beam.

Rewrite the above equation to find $\theta$.

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

Write the expression for space between the two adjacent lines.

  $d=\frac{1}{N}$        (III)

Here, $N$ is the number of lines per centimeter in the grating.

Conclusion:

Substitute $4000\text{cm}$ for $N$ in equation (III) to find $d$

  $\begin{array}{c}d=\frac{1}{4000\text{cm}}\\ =\frac{1}{\left(\text{4000}\text{cm}\right)}\\ =2.5\times {10}^{-4}\text{cm}\end{array}$

The distance between the two adjacent lines is $2.5\times {10}^{-4}\text{m}$

Substitute $2.5\times {10}^{-4}\text{m}$ for $d$, $589\text{nm}$ for $\lambda$, $1$ for $m$ in equation (II) to find $\theta$.

  $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(1\right)\left(589\times {10}^{-9}\text{m}\right)}{\left(2.5\times {10}^{-4}\text{cm}\right)\left(\frac{\text{0}\text{.01m}}{\text{1}\text{cm}}\right)}\right)\\ ={\sin }^{-1}\left(\frac{\left(589\times {10}^{-9}\text{m}\right)}{\left(2.5\times {10}^{-6}\text{cm}\right)}\right)\\ ={\sin }^{-1}\left(0.2356\right)\\ =13.6°\end{array}$

Therefore, the angular position for $m=1$ is $13.6°$

(b)

To determine

Visible lines in the grating.

Answer to Problem 26E

The number of lines can be seen in the grating is $4\text{lines}$.

Explanation of Solution

Write the expression for double slit interference.

  $\sin {\theta }_{\max }=\frac{m_{\max }\lambda }{d}$        (I)

Here, $\lambda$ is the wavelength of light, $m_{\max }$ is the maximum order of interference, $d$ is the distance between the slits and ${\theta }_{\max }$ is the maximum angle of the beam.

Rewrite the above equation to find $m_{\max }$

$m_{\max }=\frac{\sin {\theta }_{\max }\cdot d}{\lambda }$        (II)

Conclusion:

Substitute $90°$ for ${\theta }_{\max }$, $2.5\times {10}^{-4}\text{m}$ for $d$ and $589\text{nm}$ for $\lambda$ to find $m_{\max }$

  $\begin{array}{c}{m}_{\max }=\frac{\sin \left(90°\right)\cdot \left(2.5\times {10}^{-4}\text{cm}\right)\left(\frac{\text{0}\text{.01m}}{\text{1}\text{cm}}\right)}{\left(589\times {10}^{-9}\text{m}\right)}\\ =\frac{\left(1\right)\left(2.5\times {10}^{-6}\text{cm}\right)}{\left(589\times {10}^{-9}\text{m}\right)}\\ =4.24\\ \approx 4\end{array}$

Therefore, the number of lines can be seen in the grating is $4\text{lines}$.