Chapter 25, Problem 65P

(a)

To determine

The distance between the adjacent slits on the grating.

Answer to Problem 65P

The distance between the adjacent slits on the grating is $\underset{\_}{1.80\times {10}^{-6}\text{m}}$.

Explanation of Solution

Given that the number of slits per unit length of the grating is $5550\text{slits/cm}$.

Write the expression for the distance between adjacent slits.

  $d=\frac{1}{\text{number of slits per unit length}}$                                                                    (I)

Here, $d$ is the distance between adjacent slits.

Conclusion:

Substitute $5550\text{slits/cm}$ for number of slits per unit length in equation (I) to find $d$.

  $\begin{array}{c}d=\frac{1}{5550\text{slits/cm}}\\ =\frac{1\text{cm}\times \frac{1\text{m}}{100\text{cm}}}{5550\text{slits}}\\ =1.80\times {10}^{-6}\text{m}\end{array}$

Therefore, the distance between the adjacent slits on the grating is $\underset{\_}{1.80\times {10}^{-6}\text{m}}$.

(b)

To determine

The distance from the central bright spot to the first-order maximum on the screen.

Answer to Problem 65P

The distance from the central bright spot to the first-order maximum on the screen is $\underset{\_}{2.24\text{m}}$.

Explanation of Solution

Given that the wavelength of light is $0.680\text{μm}$, the distance between the grating and the screen is $5.50\text{m}$. It is obtained that the separation between the adjacent slits is $1.80\times {10}^{-6}\text{m}$.

Write the expression for the double slit maxima.

  $d\sin \theta =m\lambda m=0,\pm 1,\pm 2,\mathrm{...}$                                                                      (I)

Here, $d$ is the slit width, $\theta$ is the angle of diffraction, $m$ is the order of diffraction, and $\lambda$ is the wavelength of light.

For, first order spectrum, $m=1$, and thus equation (I) can be solved for $\theta$ as,

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

Let $x$ be the distance from the central bright spot to the first order maximum as shown in Figure 1.

From Figure 1, write the expression for $x$.

  $x=D\tan \theta$                                                                                                  (III)

Here, $D$ is the distance between the grating and the screen.

Use equation (II) in (III).

  $x=D\tan \left({\sin }^{-1}\frac{\lambda }{d}\right)$                                                                                    (IV)

Conclusion:

Substitute $5.50\text{m}$ for $D$, $0.680\text{μm}$ for $\lambda$, and $1.80\times {10}^{-6}\text{m}$ for $d$ in equation (IV) to find $x$.

  $\begin{array}{c}x=\left(5.50\text{m}\right)\tan \left({\sin }^{-1}\frac{0.680\text{μm}}{1.80\times {10}^{-6}\text{m}}\right)\\ =\left(5.50\text{m}\right)\tan \left({\sin }^{-1}\frac{0.680\text{μm}\times \frac{1\text{m}}{1\times {10}^6\text{μm}}}{1.80\times {10}^{-6}\text{m}}\right)\\ =2.24\text{m}\end{array}$

Therefore, the distance from the central bright spot to the first-order maximum on the screen is $\underset{\_}{2.24\text{m}}$.

(c)

To determine

The distance from the central bright spot to the second-order maximum on the screen.

Answer to Problem 65P

The distance from the central bright spot to the second-order maximum on the screen is $\underset{\_}{6.33\text{m}}$.

Explanation of Solution

For, first order spectrum, $m=1$, and thus equation (I) can be solved for $\theta$ as,

  $\theta ={\sin }^{-1}\frac{2\lambda }{d}$                                                                                                        (V)

Use equation (V) in (III).

  $x=D\tan \left({\sin }^{-1}\frac{2\lambda }{d}\right)$                                                                                      (VI)

Conclusion:

Substitute $5.50\text{m}$ for $D$, $0.680\text{μm}$ for $\lambda$, and $1.80\times {10}^{-6}\text{m}$ for $d$ in equation (VI) to find $x$.

  $\begin{array}{c}x=\left(5.50\text{m}\right)\tan \left({\sin }^{-1}\frac{2\times 0.680\text{μm}}{1.80\times {10}^{-6}\text{m}}\right)\\ =\left(5.50\text{m}\right)\tan \left({\sin }^{-1}\frac{2\times 0.680\text{μm}\times \frac{1\text{m}}{1\times {10}^6\text{μm}}}{1.80\times {10}^{-6}\text{m}}\right)\\ =6.33\text{m}\end{array}$

Therefore, the distance from the central bright spot to the second-order maximum on the screen is $\underset{\_}{6.33\text{m}}$.

(d)

To determine

Whether the assumption $\sin \theta \approx \tan \theta$ is valid in the given problem or not.

Answer to Problem 65P

The assumption $\sin \theta \approx \tan \theta$ is not valid in the given problem.

Explanation of Solution

The small angle approximation in trigonometry is used in physical problems, when the angles under consideration are very small. If the angles are considerably large, these approximations cannot be applied.

In the given problem, the angle of diffraction corresponding to the first order diffraction is given by equation (II).

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

Conclusion:

Substitute $0.680\text{μm}$ for $\lambda$, and $1.80\times {10}^{-6}\text{m}$ for $d$ in equation (II) to find $\theta$.

  $\begin{array}{c}\theta ={\sin }^{-1}\frac{0.680\text{μm}}{1.80\times {10}^{-6}\text{m}}\\ ={\sin }^{-1}\frac{0.680\text{μm}\times \frac{1\text{m}}{1\times {10}^6\text{μm}}}{1.80\times {10}^{-6}\text{m}}\\ =22.2°\end{array}$

For an angle $22.2°$, $\sin 22.2°=0.387$ and $\tan 22.2°=0.408$ which are differ by $8\%$.

Therefore, the assumption $\sin \theta \approx \tan \theta$ is not valid in the given problem.