Chapter 33, Problem 74P

(a)

To determine

To show that the phase difference between the waves from two adjacent slits is given by $\varphi =\frac{2\pi d}{\lambda }sin\theta$ .

Explanation of Solution

Given:

The path difference is calculated by considering two slits separated by a distance $d$ .

Formula used:

Write the equation of sinusoidal wave coming from first slit.

  $y=A\sin k\left(x+\lambda \right)$

Here, $A$ is the amplitude of sinusoidal wave, $\lambda$ is the angle of wave.

Write the path difference of a sinusoidal wave is given as

  $\Delta x=dsin\theta$ ....... (1)

Here, $d$ is the distance between the two slits.

Calculation:

A sine wave repeats itself after a regular interval of $2\pi$ . So,

  $\begin{array}{c}k\lambda =2\pi \\ k=\frac{2\pi }{\lambda }\end{array}$

Suppose two waves having same phase initially travel different distances $x$ and $x+\Delta x$ interfere at point. Then two waves can be written as

  $\begin{array}{c}{y}_1=A_1\sin \left(kx+\omega t\right)\\ y_2=A_2\sin \left(k\left(x+\Delta x\right)-\omega t\right)\\ =A_2\sin \left(kx-\omega t+\varphi \right)\end{array}$

Here, $\varphi$ is the phase difference of two waves at a point.

  $\varphi =k\Delta x$

Substitute $\frac{2\pi }{\lambda }$ for $k$ and $dsin\theta$ for $\Delta x$ in above equation.

  $\varphi =\frac{2\pi d}{\lambda }sin\theta$

Conclusion:

Thus, phase difference is given as $\varphi =\frac{2\pi d}{\lambda }sin\theta$ .

To determine

The differentiated form of phase difference is given by $d\varphi =\frac{2\pi d}{\lambda }cos\theta d\theta$ .

Explanation of Solution

Given:

The phase difference of two waves is given as

  $\varphi =\frac{2\pi d}{\lambda }sin\theta$   ........ (2)

Formula used:

Write the differentiateof $sin\theta$ ,

  $\frac{d\left(sin\theta \right)}{d\theta }=cos\theta$

Rearrange the above equation.

  $d\left(sin\theta \right)=cos\theta d\theta$

Calculation:

Differentiate the equation (1).

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

Substitute $cos\theta d\theta$ for $d\left(sin\theta \right)$ .

  $d\varphi =\frac{2\pi d}{\lambda }cos\theta d\theta$

Conclusion:

Thus,the differentiatedform of phase difference is given by $d\varphi =\frac{2\pi d}{\lambda }cos\theta d\theta$ .

To determine

The angular separation between the intensity maximum and minimum for some wavelength is $d\theta =\frac{\lambda }{Nd\text{cos}\theta }$ .

Explanation of Solution

Given:

The angular separation between an interference maximum and an interference minimum corresponds to a phase change of

  $d\varphi =\frac{2\pi }{N}$ ...... (3)

Formula Used:

Write the differentiate of phase difference is given by

  $d\varphi =\frac{2\pi d}{\lambda }cos\theta d\theta$ ....... (4)

Calculation:

Substitute $\frac{2\pi }{N}$ for $d\varphi$ in equation (4).

  $\frac{2\pi }{N}=\frac{2\pi d}{\lambda }cos\theta d\theta$

Rearrange the above equation.

  $d\theta =\frac{\lambda }{Ndcos\theta d\theta }$

Conclusion:

Thus, the angular separation between the intensity maximum and minimum for some wavelength is $d\theta =\frac{\lambda }{Nd\text{cos}\theta }$ .

(d)

To determine

The angular separation of the $m\text{th}$ order maximum for two nearly equal wavelengths differing by $d\lambda$ is given by $d\theta =\frac{md\lambda }{d\cos \theta }$ .

Explanation of Solution

Given:

The angle of $m\text{th}$ order interference maximum for wavelength $\lambda$ is given by

  $d\sin \theta =m\lambda$ ....... (5)}

Formula used:

Write the differentiation of $sin\theta$

  $\frac{d\left(sin\theta \right)}{d\theta }=cos\theta$

Rearrange the above equation.

  $d\left(sin\theta \right)=cos\theta d\theta$

Calculation:

Differentiate equation (5).

  $dd\left(sin\theta \right)=md\lambda$

Substitute $\cos \theta d\theta$ for $d\left(sin\theta \right)$ and rearrange.

  $\begin{array}{c}d\cos \theta d\theta =md\lambda \\ d\theta =\frac{md\lambda }{d\cos \theta }\end{array}$

Conclusion:

Thus,the angular separation of the $m\text{th}$ order maximum for two nearly equal wavelengths differing by $d\lambda$ is given by $d\theta =\frac{md\lambda }{d\cos \theta }$ .

To determine

The resolving power is given as

  $\begin{array}{c}R=\frac{\lambda }{\Delta \lambda }\\ =mN\end{array}$

Explanation of Solution

Given:

The angular separation of the $m\text{th}$ order maximum for two nearly equal wavelengths differing by $d\lambda$ is given by

  $d\theta =\frac{md\lambda }{d\cos \theta }$ ....... (6)

The angular separation between the intensity maximum and minimum for some wavelength is

  $d\theta =\frac{\lambda }{Nd\text{cos}\theta }$   ........(7)

Formula Used:

The resolving power of a grating is given by

  $R=\frac{\lambda }{d\lambda }$

Calculation:

Equate equation (7) and equation (8).

  $\frac{md\lambda }{d\cos \theta }=\frac{\lambda }{Nd\cos \theta }$

Rearrange the above equation.

  $\frac{\lambda }{d\lambda }=mN$

Conclusion:

Thus,the resolving power is given as $\frac{\lambda }{d\lambda }=mN$