Chapter 25, Problem 64P

(a)

To determine

The distance between the adjacent slits on the grating.

Answer to Problem 64P

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

Explanation of Solution

The distance between adjacent slits is the reciprocal of the number of slits per unit length.

Write the equation for the distance between the adjacent slits.

  $d=\frac{1}{n}$        (I)

Here, $d$ is the distance between adjacent slits and $n$ is the number of slits per unit length.

Conclusion:

Substitute $5550\text{ slits/cm}$ for $n$ in equation (I) to find $d$ .

  $\begin{array}{c}d=\frac{1}{5550\text{ slits/cm}}\\ =1.8018\times {10}^{-6}\text{ m}\\ \approx 1.80\times {10}^{-6}\text{ m}\end{array}$

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

(b)

To determine

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

Answer to Problem 64P

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

Explanation of Solution

Write the Bragg’s law.

  $d\sin \theta =m\lambda$        (II)

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

Substitute $1$ for $m$ in the above equation since it is asked to find the distance of the first order maximum and rewrite it for $\theta$ .

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

The arrangement is shown in figure 1.

Refer to figure 1 and write the expression for $\tan \theta$ .

  $\tan \theta =\frac{x}{D}$

Here, $x$ is the distance between the maximum and the central bright spot and $D$ is the distance between the grating and the screen.

Rewrite the above equation for $x$ .

  $x=D\tan \theta$        (IV)

Put equation (III) in equation (IV).

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

Conclusion:

Substitute $5.50\text{ m}$ for $D$ , $0.680\text{ μm}$ for $\lambda$ and $1.8018\times {10}^{-6}\text{ m}$ for $d$ in equation (V) to find $x$ .

  $\begin{array}{c}x=\left(5.50\text{ m}\right)\tan \left({\sin }^{-1}\frac{0.680\text{ μm}\frac{1\text{ m}}{{10}^6\text{ μm}}}{1.8018\times {10}^{-6}\text{ m}}\right)\\ =2.24\text{ m}\end{array}$

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

(c)

To determine

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

Answer to Problem 64P

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

Explanation of Solution

Substitute $1$ for $m$ in equation (II) since it is asked to find the distance of the second order maximum and rewrite it for $\theta$ .

  $\begin{array}{c}d\sin \theta =\left(2\right)\lambda \\ \sin \theta =\frac{2\lambda }{d}\\ \theta ={\sin }^{-1}\frac{2\lambda }{d}\end{array}$        (VI)

Put equation (VI) in equation (IV).

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

Conclusion:

Substitute $5.50\text{ m}$ for $D$ , $0.680\text{ μm}$ for $\lambda$ and $1.8018\times {10}^{-6}\text{ m}$ for $d$ in equation (VII) to find $x$ .

  $\begin{array}{c}x=\left(5.50\text{ m}\right)\tan \left({\sin }^{-1}\frac{2\left(0.680\text{ μm}\frac{1\text{ m}}{{10}^6\text{ μm}}\right)}{1.8018\times {10}^{-6}\text{ m}}\right)\\ =6.33\text{ m}\end{array}$

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

(d)

To determine

Whether it can be assumed that $\sin \theta \approx \tan \theta$ .

Answer to Problem 64P

It cannot be assumed that $\sin \theta \approx \tan \theta$ since the angles are not small.

Explanation of Solution

The assumption $\sin \theta \approx \tan \theta$ will be valid when $\theta$ is small.

Rewrite equation (II) for $\theta$ .

  $\begin{array}{c}d\sin \theta =m\lambda \\ \sin \theta =\frac{m\lambda }{d}\\ \theta ={\sin }^{-1}\frac{m\lambda }{d}\end{array}$        (VIII)

Conclusion:

Substitute $1$ for $m$ , $0.680\text{ μm}$ for $\lambda$ and $1.8018\times {10}^{-6}\text{ m}$ for $d$ in equation (VIII) to find $\theta$ corresponding to first order maximum.

  $\begin{array}{c}\theta ={\sin }^{-1}\frac{1\left(0.680\text{ μm}\frac{1\text{ m}}{{10}^6\text{ μm}}\right)}{1.8018\times {10}^{-6}\text{ m}}\\ =22.17°\end{array}$

Substitute $2$ for $m$ , $0.680\text{ μm}$ for $\lambda$ and $1.8018\times {10}^{-6}\text{ m}$ for $d$ in equation (VIII) to find $\theta$ corresponding to second order maximum.

  $\begin{array}{c}\theta ={\sin }^{-1}\frac{2\left(0.680\text{ μm}\frac{1\text{ m}}{{10}^6\text{ μm}}\right)}{1.8018\times {10}^{-6}\text{ m}}\\ =49.0°\end{array}$

From the calculations, it is clear that the values of $\theta$ is not small. Hence the approximation is not valid.

Therefore, it cannot be assumed that $\sin \theta \approx \tan \theta$ since the angles are not small.