Chapter 38, Problem 38.72AP

(a)

To determine

To show: The point where $I=0.5I_{\max }$ must have $\varphi =\sqrt{2}\sin \varphi$.

Answer to Problem 38.72AP

The point where $I=0.5I_{\max }$ must have $\varphi =\sqrt{2}\sin \varphi$.

Explanation of Solution

Given info: The equation of the intensity of the light in the diffraction pattern is $I=I_{\max }\frac{{\sin }^2\varphi }{{\varphi }^2}$ where $\varphi =\frac{\left(\pi a\sin \theta \right)}{\lambda }$

The formula to calculate the intensity of the light is,

$\begin{array}{c}I=I_{\max }\frac{{\sin }^2\varphi }{{\varphi }^2}\\ \frac{I}{I_{\max }}=\frac{{\sin }^2\varphi }{{\varphi }^2}\end{array}$

Here,

$I_{\max }$ is the maximum intensity of the light.

$\varphi$ is the phase constant of the light.

The value of $I=0.5I_{\max }$.

Substitute $0.5I_{\max }$ for $I$ in above equation to find the value of $\varphi$.

$\begin{array}{c}\frac{0.5I_{\max }}{I_{\max }}=\frac{{\sin }^2\varphi }{{\varphi }^2}\\ \frac{{\sin }^2\varphi }{{\varphi }^2}=\frac{1}{2}\\ \sin \varphi =\frac{\varphi }{\sqrt{2}}\\ \varphi =\sqrt{2}\sin \varphi \end{array}$

Conclusion

Therefore, the point where $I=0.5I_{\max }$ must have $\varphi =\sqrt{2}\sin \varphi$.

(b)

To determine

To draw: Plot $y_1=\sin \varphi$ and $y_2=\frac{\varphi }{\sqrt{2}}$ on the same set of axes over a range from $\varphi =1\text{rad}$ to $\varphi =\frac{\pi }{2}\text{rad}$.

Answer to Problem 38.72AP

The graph between $y_1=\sin \varphi$ and $y_2=\frac{\varphi }{\sqrt{2}}$ on the same set of axes over a range from $\varphi =1\text{rad}$ to $\varphi =\frac{\pi }{2}\text{rad}$.

Figure (1)

Explanation of Solution

Given info: The equation of the intensity of the light in the diffraction pattern is $I=I_{\max }\frac{{\sin }^2\varphi }{{\varphi }^2}$ where $\varphi =\frac{\left(\pi a\sin \theta \right)}{\lambda }$

The equation of $y_1$ is $\sin \varphi$ and the equation for $y_2$ is $\frac{\varphi }{\sqrt{2}}$ over a range from $\varphi =1\text{rad}$ to $\varphi =\frac{\pi }{2}\text{rad}$ is shown in the figure below.

The solution of both the equation to coincide at a point is ,

$\begin{array}{c}\sin \varphi =\frac{\varphi }{\sqrt{2}}\\ \varphi =1.39\text{rad}\end{array}$

So the solution of the transcendental equation is $\varphi =1.39\text{rad}$.

(c)

To determine

To show: The angular full width at half maximum of the central diffraction maximum is $\theta =\frac{0.885\lambda }{a}$ if the fraction $\frac{\lambda }{a}$ is not large.

Answer to Problem 38.72AP

The angular full width at half maximum of the central diffraction maximum is $\theta =\frac{0.885\lambda }{a}$ if the fraction $\frac{\lambda }{a}$ is not large.

Explanation of Solution

Given info: The equation of the intensity of the light in the diffraction pattern is $I=I_{\max }\frac{{\sin }^2\varphi }{{\varphi }^2}$ where $\varphi =\frac{\left(\pi a\sin \theta \right)}{\lambda }$

The formula to calculate the phase angle is,

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

Rewrite the above equation for $\sin \theta$.

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

If the value of $\frac{\lambda }{a}$ is small then,

$\theta \approx \left(\frac{\varphi }{\pi }\right)\frac{\lambda }{a}$

The path covered by the light is symmetric so the phase angle is double the initial value.

$\theta =2\left(\frac{\varphi }{\pi }\right)\frac{\lambda }{a}$

Substitute $1.39\text{rad}$ for $\varphi$ in above equation to find the value of $\theta$.

$\begin{array}{c}\theta =2\left(\frac{1.39\text{rad}}{3.14\text{rad}}\right)\frac{\lambda }{a}\\ =\frac{0.885\lambda }{a}\end{array}$

Conclusion

Therefore, the angular full width at half maximum of the central diffraction maximum is $\theta =\frac{0.885\lambda }{a}$ if the fraction $\frac{\lambda }{a}$ is not large.

(d)

To determine

The number of steps involved to solve the transcendental equation $\varphi =\sqrt{2}\sin \varphi$.

Answer to Problem 38.72AP

The number of steps involved to solve the transcendental equation $\varphi =\sqrt{2}\sin \varphi$ is around $13$.

Explanation of Solution

Given info: The equation of the intensity of the light in the diffraction pattern is $I=I_{\max }\frac{{\sin }^2\varphi }{{\varphi }^2}$ where $\varphi =\frac{\left(\pi a\sin \theta \right)}{\lambda }$

The equation of $y_1$ is $\sqrt{2}\sin \varphi$ and the equation for $y_2$ is $\varphi$, the value of $\varphi$ is taken from $1$ to $2$ to find the solution of the equation and the corresponding values are shown in the table below.

$\varphi$	$\sqrt{2}\sin \varphi$
$1$	$1.19$
$2$	$1.29$
$1.5$	$1.41$
$1.4$	$1.394$
$1.39$	$1.391$
$1.392$	$1.3917$
$1.3915$	$1.39154$
$1.39152$	$1.39155$
$1.3916$	$1.39158$
$1.39158$	$1.391563$
$1.39157$	$1.391561$
$1.39156$	$1.391558$
$1.3915574$	$1.3915574$

The solution of the transcendental equation $\varphi =\sqrt{2}\sin \varphi$ is $1.3915574$ and the solution can be achieved in around $13$ steps.

Conclusion

Therefore, the number of steps involved to solve the transcendental equation $\varphi =\sqrt{2}\sin \varphi$ is around $13$.