Chapter 36, Problem 36PQ

(a)

To determine

The angle at which the third order maximum occur.

Answer to Problem 36PQ

The angle at which the third order maximum occur is $43.85°$.

Explanation of Solution

Write the expression for $m^{\text{th}}$ maxima of diffraction grating.

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

Here, $d$ is the spacing grating, $\lambda$ is the wavelength and $m$ is the order.

Conclusion:

Calculate the grating spacing as follows.

    $\begin{array}{c}d=\frac{1}{3.65\times {10}^3\text{rulings}/\text{cm}}\\ =2.74\times {10}^{-4}\text{cm}\times \frac{1\text{m}}{100\text{cm}}\\ =2.74\times {10}^{-6}\text{m}\end{array}$

Substitute $3$ for $m$, $632.8\text{nm}$ for $\lambda$ and $2.74\times {10}^{-6}\text{m}$ for $d$ in equation (I) to calculate $\theta$.

    $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{3\times \left(632.8\text{nm}\times \frac{1\text{m}}{{10}^9\text{nm}}\right)}{2.74\times {10}^{-6}\text{m}}\right)\\ ={\sin }^{-1}\left(0.6928\right)\\ =43.85°\end{array}$

Therefore, the angle at which the third order maximum occur is $43.85°$.

(b)

To determine

The angle at which third order maximum occur if the experiment was carried out underwater.

Answer to Problem 36PQ

The angle at which third order maximum occur if the experiment was carried out underwater is $31.34°$.

Explanation of Solution

Write the expression to calculate wavelength of light in water.

    ${\lambda }_{\text{water}}=\frac{\lambda }{n}$                                                                                                                  (II)

Here, ${\lambda }_{\text{water}}$ is the wavelength of light in water, $\lambda$ is the wavelength of light in air and $n$ is the refraction index of water.

Write the expression for $m^{\text{th}}$ maxima of diffraction grating.

    $\begin{array}{l}d\sin \theta =m{\lambda }_{\text{water}}\\ \sin \theta =\frac{m{\lambda }_{\text{water}}}{d}\\ \theta ={\sin }^{-1}\left(\frac{m{\lambda }_{\text{water}}}{d}\right)\end{array}$                                                                                                (III)

Here, $d$ is the spacing grating, ${\lambda }_{\text{water}}$ is the wavelength in water and $m$ is the order.

Conclusion:

Substitute $632.8\text{nm}$ for $\lambda$ and $1.33$ for $n$ in equation (II) to calculate ${\lambda }_{\text{water}}$.

    $\begin{array}{c}{\lambda }_{\text{water}}=\frac{632.8\text{nm}}{1.33}\\ =475.79\text{nm}\end{array}$

Substitute $3$ for $m$, $475.79\text{nm}$ for $\lambda$ and $2.74\times {10}^{-6}\text{m}$ for $d$ in equation (III) to calculate $\theta$.

    $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{3\times \left(475.79\text{nm}\times \frac{1\text{m}}{{10}^9\text{nm}}\right)}{2.74\times {10}^{-6}\text{m}}\right)\\ ={\sin }^{-1}\left(0.52\right)\\ =31.34°\end{array}$

Therefore, the angle at which third order maximum occur if the experiment was carried out underwater is $31.34°$.

(c)

To determine

The relationship between the diffracted rays from part (a) and part (b).

Answer to Problem 36PQ

The relationship between the diffracted rays from part (a) and part (b) is ${\theta }_{\text{water}}={\sin }^{-1}\left(\frac{m\lambda }{d\cdot n}\right)$.

Explanation of Solution

Write the expression for $m^{\text{th}}$ maxima of diffraction grating.

    $\begin{array}{l}d\sin {\theta }_{\text{water}}=m{\lambda }_{\text{water}}\\ \sin {\theta }_{\text{water}}=\frac{m{\lambda }_{\text{water}}}{d}\\ {\theta }_{\text{water}}={\sin }^{-1}\left(\frac{m{\lambda }_{\text{water}}}{d}\right)\end{array}$                                                                                             (IV)

Here, $d$ is the spacing grating, ${\lambda }_{\text{water}}$ is the wavelength in water and m is the order.

Write the expression for wavelength of light in water.

    ${\lambda }_{\text{water}}=\frac{\lambda }{n}$

Here, $\lambda$ is the water and $n$ is the refraction index of water.

Substitute $\left(\frac{\lambda }{n}\right)$ for ${\lambda }_{\text{water}}$ in equation (IV) to calculate ${\theta }_{\text{water}}$.

    ${\theta }_{\text{water}}={\sin }^{-1}\left(\frac{m\lambda }{d\cdot n}\right)$

Conclusion:

Therefore, the relationship between the diffracted rays from part (a) and part (b) is ${\theta }_{\text{water}}={\sin }^{-1}\left(\frac{m\lambda }{d\cdot n}\right)$.