Chapter 27, Problem 5TP

To determine

The measurement which remain unchanged after the amplitude of two waves are doubled.

Answer to Problem 5TP

The intensity of dark fringe.remains unchanged. Option (b) is correct.

Explanation of Solution

Introduction:

The resultant amplitude of two waves with equal amplitude is given by

  $A^2=2A_1{}^2\left(1+\cos \delta \right)$

The relation between intensity and amplitude is given by

  $\begin{array}{c}I\propto A^2\\ I=k{A}^2\end{array}$

The difference in intensity of consecutive bright and dark fringes is given by

  $I_d=I_{Max}-I_{Min}$

The double of amplitude is given by

  $A^{\prime }=2A$

The maximum amplitude of interference is calculated as follows:

  $\begin{array}{c}{A}_{Max}{}^2=2A_1{}^2\left(1+\cos \delta \right)\\ =2A_1{}^2\left(1+\cos 0\right)\\ =4A_1{}^2\end{array}$

The maximum intensity of interference is calculated as follows:

  $\begin{array}{c}{I}_{Max}=k{A}_{Max}{}^2\\ =k\left(4A_1{}^2\right)\\ =4k{A}_1{}^2\end{array}$

The minimum amplitude of interference is calculated as follows:

  $\begin{array}{c}{A}_{Mini}{}^2=2A_1{}^2\left(1+\cos \delta \right)\\ =2A_1{}^2\left(1+\cos \pi \right)\\ =2A_1{}^2\left(0\right)\\ =0\end{array}$

The minimum intensity of interference is calculated as follows:

  $\begin{array}{c}{I}_{Mini}=k{A}_{Mini}{}^2\\ =k\left(0\right)\\ =0\end{array}$

The difference in intensity of consecutive bright and dark fringes is calculated as follows:

  $\begin{array}{c}{I}_d=I_{Max}-I_{Min}\\ =4k{A}_1{}^2-0\\ =4k{A}_1{}^2\end{array}$

The double of amplitude is calculated as follows:

  $A_1{}^{\prime }=2A_1$

The maximum amplitude of interference having doubled amplitude waves is calculated as follows:

  $\begin{array}{c}{\left(A_{Max}{}^{\prime }\right)}^2=2{\left(A_1{}^{\prime }\right)}^2\left(1+\cos \delta \right)\\ =2{\left(2A_1\right)}^2\left(1+\cos 0\right)\\ =16A_1{}^2\end{array}$

The maximum intensity of interference having doubled amplitude waves is calculated as follows:

  $\begin{array}{c}{I}_{Max}{}^{\prime }=k{\left(A_{Max}{}^{\prime }\right)}^2\\ =k\left(16A_1{}^2\right)\\ =16k{A}_1{}^2\end{array}$

The minimum amplitude of interference having doubled amplitude waves is calculated as follows:

  $\begin{array}{c}{\left(A_{Mini}{}^{\prime }\right)}^2=2{\left(A_1{}^{\prime }\right)}^2\left(1+\cos \delta \right)\\ =2{\left(2A_1\right)}^2\left(1+\cos \pi \right)\\ =8A_1{}^2\left(0\right)\\ =0\end{array}$

The minimum intensity of interference having doubled amplitude waves is calculated as follows:

  $\begin{array}{c}{I}_{Mini}{}^{\prime }=k{\left(A_{Mini}{}^{\prime }\right)}^2\\ =k\left(0\right)\\ =0\end{array}$

The difference in intensity of consecutive bright and dark fringes having doubled amplitude waves is calculated as follows:

  $\begin{array}{c}{I}_d{}^{\prime }=I_{Max}{}^{\prime }-I_{Min}{}^{\prime }\\ =16k{A}_1{}^2-0\\ =16k{A}_1{}^2\end{array}$

Conclusion:

The intensity of dark fringe remains unchanged as $0$. Therefore option (b) is correct.

The intensity of bright fringe is changed to $16k{A}_1{}^2$ form $4k{A}_1{}^2$. Therefore option (a) is incorrect.

The difference in intensitities of consecutive bright and dark fringes is changed to $16k{A}_1{}^2$ form $4k{A}_1{}^2$. Therefore option (c) is incorrect.

The intensity of dark fringe remains unchanged. So one potion is correct Therefore option (d) is incorrect.