Chapter 38, Problem 77CP

(a)

To determine

The angle for the first minimum in the diffraction pattern.

Answer to Problem 77CP

The angle for the first minimum in the diffraction pattern is $41.8°$ .

Explanation of Solution

Write the expression for the minima.

    $a\sin \theta =m\lambda$

Here, $a$ is the slit width, $\theta$ is the angle, $m$ is the order, and $\lambda$ is the wave length of wave.

Rewrite the above equation.

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

Write the expression to calculate the wavelength.

    $\lambda =\frac{c}{f}$                                                                                                        (II)

Here, $c$ is the speed of light and $f$ is the frequency of the wave.

Conclusion:

Substitute $7.50\text{GHz}$ for $f$ and $3.00\times {10}^8\text{ m/s}$ for $c$ in (II) to find $\lambda$.

    $\begin{array}{c}\lambda =\frac{3.00\times {10}^8\text{ m/s}}{7.50\text{GHz}}\\ =\frac{3.00\times {10}^8\text{ m/s}}{7.50\text{GHz}\times \frac{{10}^9\text{ Hz}}{1\text{GHz}}}\\ =4.00\times {10}^{-2}\text{ m}\end{array}$

Substitute $4.00\times {10}^{-2}\text{ m}$ for $\lambda$ and $6.00\text{cm}$ for $a$ in (I) to find $\theta$.

    $\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{4.00\times {10}^{-2}\text{ m}}{6.00\text{cm}}\right)\\ ={\sin }^{-1}\left(\frac{4.00\times {10}^{-2}\text{ m}}{6.00\text{cm}\times \frac{{10}^{-2}\text{ m}}{1\text{cm}}}\right)\\ =41.8°\end{array}$

Therefore, the angle for the first minimum in the diffraction pattern is $41.8°$

(b)

To determine

The relative intensity at $15.0°$ .

Answer to Problem 77CP

The relative intensity at $15.0°$ is $0.592$.

Explanation of Solution

Write the expression for the relative intensity.

    $\frac{I}{I_{\max }}={\left[\frac{\sin \left(\varphi \right)}{\varphi }\right]}^2$                                                                                         (III)

Here, $\frac{I}{I_{\max }}$ is the relative intensity and $\varphi$ is the phase angle.

Write the expression for the phase angle.

    $\varphi =\frac{\pi a\sin \theta }{\lambda }$                                                                                                  (IV)

Conclusion:

Substitute $4.00\times {10}^{-2}\text{ m}$ for $\lambda$, $41.8°$ for $\theta$ , and $6.00\text{cm}$ for $a$ in (IV) to find $\varphi$.

    $\begin{array}{c}\varphi =\frac{\pi \left(\text{6}\text{.00 cm}\right)\sin 15.0°}{4.00\times {10}^{-2}\text{ m}}\\ =\frac{\pi \left(\text{6}\text{.00 cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\sin 15.0°}{0.0400\text{ m}}\\ =1.22\text{ rad}\end{array}$

Substitute $1.22\text{ rad}$ for $\varphi$ in (III) to find $\frac{I}{I_{\max }}$.

  $\begin{array}{c}\frac{I}{I_{\max }}={\left[\frac{\sin \left(1.22\text{ rad}\right)}{1.22\text{ rad}}\right]}^2\\ =0.592\end{array}$

Therefore, the relative intensity at $15.0°$ is $0.592$ .

(c)

To determine

Maximum distance between the plane of the sources and the slit if the diffraction pattern are to be resolved.

Answer to Problem 77CP

Maximum distance between the plane of the sources and the slit if the diffraction pattern are to be resolved is $0.262\text{m}$ .

Explanation of Solution

Write the expression for the distance between the plane of the sources and the slit.

    $L=\mathcal{l}\cot \alpha$                                                                                                      (V)

Here, $L$ is the distance between the plane of the sources and the slit, $\mathcal{l}$ is the distance of one source from the midpoint of the plane between the two sources, and $\alpha$ is the angle.

Write the expression between $l$ and the distance between the sources.

    $d=2l$

Here, $d$ is the distance between the sources.

From the figure (I) $\alpha =\frac{\theta }{2}$.

Use $\frac{\theta }{2}$ for $\alpha$ and $2l$ for $d$ in (V) and rewrite.

    $L=\left(\frac{d}{2}\right)\cot \left(\frac{\theta }{2}\right)$                                                                                          (VI)

Conclusion:

Substitute $20.0\text{cm}$ for $d$ and $41.8°$ for $\theta$  in (VI) to find $L$.

    $\begin{array}{c}L=\left(\frac{20.0\text{cm}}{2}\right)\cot \left(\frac{41.8°}{2}\right)\\ =\left(\frac{20.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}}{2}\right)\cot \left(\frac{41.8°}{2}\right)\\ =0.262\text{ m}\end{array}$

Therefore, the maximum distance between the plane of the sources and the slit if the diffraction pattern are to be resolved is $0.262\text{m}$ .