Chapter 24, Problem 66GP

To determine

The angles for maximum and minimum singles.

Answer to Problem 66GP

The minimum and maximum angles are $11°,°33°,67°,113°,147°\&169°$ and $22°,47°,133°\&158°$ respectively.

Explanation of Solution

Given data:

Distance, $\text{d}=8.00\text{m}$

Frequency, $\text{f}=102.1\times {10}^6\text{Hz}$

The speed of the light, $c=3.00\times {10}^8\text{m/s}$

Formula used:

  $d\sin \theta =m\lambda$ and $\lambda =\frac{c}{\text{f}}$

Where,

  $d$ is the distance.

  $\lambda$ is wavelength.

  $\text{f}$ is the frequency.

Calculation:

The wavelength of the signal: $\lambda =\frac{c}{\text{f}}$

  $\lambda =\frac{c}{\text{f}}=\frac{3\times {10}^8\text{m/s}}{102.1\times {10}^6\text{Hz}}=2.94\text{m}$

For constructive interference: $d\sin \theta =m\lambda$

  $d\sin \theta =m\lambda ,m=0,1,2,3,\mathrm{...},$

  $\left(8.0\right)\sin {\theta }_{1\max }=\left(1\right)\left(2.94\text{m}\right)$

  $\therefore {\theta }_{1\max }=22°$

Similarly put $\text{m}=\text{2,}$

  $\therefore {\theta }_{2\max }=47°$

But $\text{m}=3$

  $\left(8.0\right)\sin {\theta }_{3\max }=3\left(2.94\text{m}\right)$

  $\therefore \sin {\theta }_{3\max }>1$

Hence, there is no third maximum.

Therefore, the angles for maxima are $22°,47°,133°\&158°$ .

For destructive interference, $d\sin \theta =\left(m-\frac{1}{2}\right)\lambda ,\text{m}=1,2,3,\mathrm{...},$

  $\left(8.0\right)\sin {\theta }_{1\min }=\left(1-\frac{1}{2}\right)\left(2.94\text{m}\right)$

  $\therefore {\theta }_{1\min }=11°$

Similarly put $\text{m}=\text{2,3}$

  $\therefore {\theta }_{2\min }=33°$

  $\therefore {\theta }_{3\min }=67°$

But $\text{m}=4$

  $\left(8.0\right)\sin {\theta }_{4\max }=\left(4-\frac{1}{2}\right)\left(2.94\text{m}\right)$

  $\therefore \sin {\theta }_{4\max }>1$

Hence, there is no fourth minimum

Therefore, the angles for minima are $11°,°33°,67°,113°,147°\&169°$ .

Conclusion: The angles of minima and maxima are $11°,°33°,67°,113°,147°\&169°$ and $22°,47°,133°\&158°$ respectively.