Chapter 38, Problem 38.77CP

Suppose the single slit in Figure 38.4 is 6.00 cm wide and in front of a microwave source operating at 7.50 GHz. (a) Calculate the angle for the first minimum in the diffraction pattern. (b) What is the relative intensity l/l_max at 0 =- 15.0°? (c) Assume two such sources, separated laterally by 20.0 cm, are behind the slit. What must be the maximum distance between the plane of the sources and the slit if the diffraction pat­terns are to be resolved? In this case, the approximation sin θ ≈ tan 8 is not valid because of the relatively small value of a/λ.

To determine

The angle for the first minimum in the diffraction pattern.

Answer to Problem 38.77CP

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

Explanation of Solution

Given info: The width of the slit is $6.00\text{cm}$ and the microwave is operating at $7.50\text{GHz}$.

The expression of wavelength ( $\lambda$) of the microwave is,

$\lambda =\frac{c}{f}$

Here,

$c$ is the speed of light.

$f$ is the frequency of microwave.

Substitute $7.50\text{GHz}$ for $f$ and $3\times {10}^8\text{m}/\text{s}$ in the above equation.

$\begin{array}{c}\lambda =\frac{3\times {10}^8\text{m}/\text{s}}{7.50\text{GHz}\left(\frac{{10}^9\text{Hz}}{1\text{GHz}}\right)}\\ =\frac{3\times {10}^8\text{m}/\text{s}}{7.50\times {10}^9\text{Hz}}\\ \lambda =0.04\text{m}\end{array}$

Thus, the wavelength of the microwave source is $0.04\text{m}$.

The expression of the condition for the first minimum in the diffraction pattern is,

$\sin \theta =\frac{m\lambda }{a}$

Here,

$m$ is the number of minimum.

$\lambda$ is the wavelength of the microwave source.

$a$ is the width of the slit.

Substitute $1$ for $m$, $6.00\text{cm}$ for $a$ and $0.04\text{m}$ for $\lambda$ in the above equation.

$\begin{array}{c}\sin \theta =\frac{1\left(0.04\text{m}\right)}{6.00\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\\ \sin \theta =0.666\\ \theta ={\sin }^{-1}\left(0.666\right)\\ \theta \approx 41.8°\end{array}$

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

Conclusion:

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

To determine

The relative intensity at $15.0°$.

Answer to Problem 38.77CP

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

Explanation of Solution

Given info: The width of the slit is $6.00\text{cm}$, the angle to find the relative intensity is $15.0°$ and the microwave is operating at $7.50\text{GHz}$.

The expression of the intensity variation in a diffraction pattern from a single slit is,

$I=I_{\max }{\left[\frac{\sin \left(\frac{\pi a\sin \theta }{\lambda }\right)}{\frac{\pi a\sin \theta }{\lambda }}\right]}^2$

Here,

$I_{\max }$ is the intensity at central maximum.

$\lambda$ is the wavelength of the microwave source.

Rearrange the above equation for $\frac{I}{I_{\max }}$.

$\begin{array}{c}I=I_{\max }{\left[\frac{\sin \left(\frac{\pi a\sin \theta }{\lambda }\right)}{\frac{\pi a\sin \theta }{\lambda }}\right]}^2\\ \frac{I}{I_{\max }}={\left[\frac{\sin \left(\frac{\pi a\sin \theta }{\lambda }\right)}{\frac{\pi a\sin \theta }{\lambda }}\right]}^2\end{array}$

Substitute $15.0°$ for $\theta$, $6.00\text{cm}$ for $a$ and $0.04\text{m}$ for $\lambda$ in the above equation.

$\begin{array}{c}\frac{I}{I_{\max }}={\left[\frac{\sin \left(\frac{\pi \times 6.00\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\sin 15.0°}{0.04\text{m}}\right)}{\frac{\pi \times 6.00\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\sin 15.0°}{0.04\text{m}}}\right]}^2\\ ={\left[\frac{\sin \left(\frac{\pi \times 0.06\text{m}\left(0.2588\right)}{0.04\text{m}}\right)}{\frac{\pi \times 0.06\text{m}\left(0.2588\right)}{0.04\text{m}}}\right]}^2\\ \frac{I}{I_{\max }}=0.592\end{array}$

Thus, the relative intensity $\frac{I}{I_{\max }}$ at $15.0°$ is $0.592$.

Conclusion

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

To determine

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

Answer to Problem 38.77CP

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

Explanation of Solution

Given info: The width of the slit is $6.00\text{cm}$, the microwave is operating at $7.50\text{GHz}$ and the distance between the two sources is $20.0\text{cm}$.

The figure1 shows the given condition.

Figure (1)

Consider $\beta$ be the angle halfway between the two sources.

The expression of distance ( $l$) of each source from the central line is,

$l=\frac{D}{2}$

Here,

$D$ is the distance between the two sources.

Substitute $20.0\text{cm}$ for $D$ in the above equation.

$\begin{array}{c}l=\frac{20.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}{2}\\ l=0.100\text{m}\end{array}$

Thus, the distance of each source from the central line is $0.100\text{m}$.

From figure1 the expression of distance ( $L$) between the plane of sources and the slit is,

$L=l\cot \beta$

The value of angle $\beta$ is equal to angle $\frac{\theta }{2}$ according to vertically opposite angle.

Substitute $\frac{\theta }{2}$ for $\beta$ in the above equation.

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

Substitute $41.8°$ for $\theta$ and $0.100\text{m}$ for $l$ in the above equation.

$\begin{array}{c}L=\left(0.100\text{m}\right)\cot \left(\frac{41.8°}{2}\right)\\ =\left(0.100\text{m}\right)\cot \left(20.9°\right)\\ =0.2618\text{m}\\ =0.262\text{m}\end{array}$

Conclusion:

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