Arrange the resultant intensities in increasing order, where two interfering waves of equal intensities have following path differences:
A. \( 0 \)
B. \( \lambda/4 \)
C. \( \lambda/6 \)
D. \( \lambda/3 \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → D, B, C, A
For two waves of equal intensity \( I_0 \), the resultant intensity depends on the phase difference, and the phase difference follows from the path difference:
\( \displaystyle \phi = \frac{2\pi}{\lambda}\,\Delta x \)
\( \displaystyle I = 4I_0\cos^{2}\frac{\phi}{2} \)
\( \text{A: } \Delta x = 0,\ \phi = 0,\ I = 4I_0 \)
\( \text{B: } \Delta x = \lambda/4,\ \phi = \pi/2,\ I = 4I_0\cos^{2}45^\circ = 2I_0 \)
\( \text{C: } \Delta x = \lambda/6,\ \phi = \pi/3,\ I = 4I_0\cos^{2}30^\circ = 3I_0 \)
\( \text{D: } \Delta x = \lambda/3,\ \phi = 2\pi/3,\ I = 4I_0\cos^{2}60^\circ = I_0 \)
Increasing order: D (\( I_0 \)) < B (\( 2I_0 \)) < C (\( 3I_0 \)) < A (\( 4I_0 \)), i.e. D, B, C, A.
Shortcut: Between 0 and \( \lambda/2 \) the intensity falls steadily as the path difference grows, so simply order the path differences: \( 0 \lt \lambda/6 \lt \lambda/4 \lt \lambda/3 \) gives intensities decreasing as A, C, B, D; reversed, D, B, C, A.