Chapter 35, Problem 52PQ

(a)

To determine

The intensity at given angular positions from the fringe.

Answer to Problem 52PQ

The intensity at $20°$ is $4.55\text{W}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Write the expression for angular location of particular maximum.

    $\alpha =\frac{\pi d}{\lambda }\sin \theta$                                             (I)

Here, $\lambda$ is wavelength of sound, $d$ is width of the slits, and $\theta$ is the angle of diffraction for a particular minimum.

Write the expression for intensity at any point on the screen due to single slit diffraction.

    $I=I_{\max }{\left(\frac{\sin \alpha }{\alpha }\right)}^2$                                                        (II)

Here, $I$ is intensity at any point, $I_{\max }$ is maximum intensity corresponding to central maximum, and $\alpha$ is angular location of particular maximum.

Conclusion:

Substitute $20°$ for $\theta$, $426\text{nm}$ for $\lambda$, and $4.5\mu \text{m}$ for $d$ in equation (I) to calculate $\alpha$.

    $\begin{array}{c}\alpha =\frac{\pi \times 4.5\mu \text{m}\times \frac{{10}^{-6}\text{m}}{1\mu \text{m}}}{426\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}}\sin \left(20°\times \frac{\pi }{180}\right)\\ =\frac{\pi \times 4.5\mu \text{m}\times \frac{{10}^{-6}\text{m}}{1\mu \text{m}}}{426\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}}\sin \left(0.523\text{radian}\right)\\ =11.33\text{radian}\end{array}$

Substitute $11.33\text{radian}$ for $\alpha$, and $655\text{W}/{\text{m}}^{\text{2}}$ for $I_{\max }$ in equation (II) to calculate $I$.

    $\begin{array}{c}I=655\text{W}/{\text{m}}^{\text{2}}{\left(\frac{\sin \left(11.33\text{radian}\right)}{11.33\text{radian}}\right)}^2\\ =655\text{W}/{\text{m}}^{\text{2}}\times {\left(\frac{0.9445}{11.33\text{radian}}\right)}^2\\ =655\text{W}/{\text{m}}^{\text{2}}\times 6.9508\times {10}^{-3}\\ =4.55\text{W}/{\text{m}}^{\text{2}}\end{array}$

Therefore, the intensity at $20°$ is $4.55\text{W}/{\text{m}}^{\text{2}}$.

(b)

To determine

The intensity at given angular positions from the fringe.

Answer to Problem 52PQ

The intensity at $40°$ is $0.572\text{W}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Conclusion:

Substitute $40°$ for $\theta$, $426\text{nm}$ for $\lambda$, and $4.5\mu \text{m}$ for $d$ in equation (I) to calculate $\alpha$.

    $\begin{array}{c}\alpha =\frac{\pi \times 4.5\mu \text{m}\times \frac{{10}^{-6}\text{m}}{1\mu \text{m}}}{426\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}}\sin \left(40°\times \frac{\pi }{180}\right)\\ =\frac{\pi \times 4.5\mu \text{m}\times \frac{{10}^{-6}\text{m}}{1\mu \text{m}}}{426\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}}\sin \left(0.698\text{radian}\right)\\ =21.31\text{radian}\end{array}$

Substitute $21.31\text{radian}$ for $\alpha$, and $655\text{W}/{\text{m}}^{\text{2}}$ for $I_{\max }$ in equation (II) to calculate $I$.

    $\begin{array}{c}I=655\text{W}/{\text{m}}^{\text{2}}{\left(\frac{\sin \left(21.31\text{radian}\right)}{21.31\text{radian}}\right)}^2\\ =655\text{W}/{\text{m}}^{\text{2}}\times {\left(\frac{0.6296}{21.31\text{radian}}\right)}^2\\ =655\text{W}/{\text{m}}^{\text{2}}\times 8.7313\times {10}^{-4}\\ =0.572\text{W}/{\text{m}}^{\text{2}}\end{array}$

Therefore, the intensity at $40°$ is $0.572\text{W}/{\text{m}}^{\text{2}}$.

(c)

To determine

The intensity at given angular positions from the fringe.

Answer to Problem 52PQ

The intensity at $60°$ is $0.15\text{W}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Conclusion:

Substitute $60°$ for $\theta$, $426\text{nm}$ for $\lambda$, and $4.5\mu \text{m}$ for $d$ in equation (I) to calculate $\alpha$.

    $\begin{array}{c}\alpha =\frac{\pi \times 4.5\mu \text{m}\times \frac{{10}^{-6}\text{m}}{1\mu \text{m}}}{426\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}}\sin \left(60°\times \frac{\pi }{180}\right)\\ =\frac{\pi \times 4.5\mu \text{m}\times \frac{{10}^{-6}\text{m}}{1\mu \text{m}}}{426\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}}\sin \left(1.047\text{radian}\right)\\ =28.72\text{radian}\end{array}$

Substitute $28.72\text{radian}$ for $\alpha$, and $655\text{W}/{\text{m}}^{\text{2}}$ for $I_{\max }$ in equation (II) to calculate $I$.

    $\begin{array}{c}I=655\text{W}/{\text{m}}^{\text{2}}{\left(\frac{\sin \left(28.72\text{radian}\right)}{28.72\text{radian}}\right)}^2\\ =655\text{W}/{\text{m}}^{\text{2}}\times {\left(\frac{-0.4310}{28.72\text{radian}}\right)}^2\\ =655\text{W}/{\text{m}}^{\text{2}}\times 2.253\times {10}^{-4}\\ =0.15\text{W}/{\text{m}}^{\text{2}}\end{array}$

Therefore, the intensity at $60°$ is $0.15\text{W}/{\text{m}}^{\text{2}}$.