Question

Two beams of light having intensities $I$ and \[4I\] interfere to produce a fringe pattern on a screen. If the phase difference between the beams is $\dfrac{\pi }{2}$ at point A and $\pi $ at point B then the difference between the resultant intensities at A and B is:
A. \[4I\]
B. \[2I\]
C. $5I$
D. $7I$

Answer 1

Hint: The intensity of the light wave is the phenomenon from which we can know the brightness of the lights. The formula of the resultant intensity of the two intensities of the interfered lights has to be known. The resultant intensities are to be calculated for two cases putting the two given phase differences. The phase differences are given for the two points. In the problem, the difference between these two resultant intensities is asked.

Formula used:
For the two intensities ${I_1}$ and ${I_2}$, the resultant intensity will be $I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \varphi $
$\varphi $ is the phase difference.

Complete step by step solution:
When Two beams of light having intensities ${I_1}$ and ${I_2}$ interfere and produce a fringe pattern on a screen,
the resultant intensity, $I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \varphi $, $\varphi $ is the phase difference.
For point A, two beams of light having intensities $I$ and \[4I\] interfere,
Hence, the resultant intensity will be ${I_A} = I + 4I + 2\sqrt {I.4I} \cos \varphi $
The phase difference at point A, $\varphi = \dfrac{\pi }{2}$
$\therefore {I_A} = 5I + 2\sqrt {4{I^2}} \cos \dfrac{\pi }{2}$
$ \Rightarrow {I_A} = 5I + 4I \times 0$ [since, $\cos \dfrac{\pi }{2} = 0$ ]
$ \Rightarrow {I_A} = 5I$
For point B, two beams of light having intensities $I$ and \[4I\] interfere
Hence, the resultant intensity will be ${I_B} = I + 4I + 2\sqrt {I.4I} \cos \varphi $
The phase difference at point B, $\varphi = \pi $
\[\therefore {I_B} = 5I + 2\sqrt {4{I^2}} \cos \pi \]
$ \Rightarrow {I_B} = 5I + 4I( - 1)$ [since, $\cos \pi = - 1$ ]
$ \Rightarrow {I_B} = 5I - 4I$
$ \Rightarrow {I_B} = I$
Therefore, the difference between the two resultant intensities is
${I_A} - {I_B} = 5I - I = 4I$
Hence, the right answer is in option (A).

Note:
The amplitudes of these interfered lights are directly proportional to the square roots of the intensities. That means, the amplitude$A \propto \sqrt I $
So, for the above problem,
For points A and B the ratio of the amplitudes is,
$\dfrac{{{A_A}}}{{{A_B}}} = \dfrac{{\sqrt {5I} }}{{\sqrt {2I} }}$
$ \Rightarrow \dfrac{{{A_A}}}{{{A_B}}} = \sqrt {\dfrac{5}{2}} $