Q149 / 180 · COMEDK 2025 (25 May Shift) · Physics
In a Young's double slit experiment, the two interfering waves forming the interference pattern have the same amplitude. The phase difference between the two waves reaching a point X is $\dfrac{\pi}{3}$. What will be the phase difference between the two waves at a point Y if the intensity ratio at X and Y is $3: 2$ ?
- A$\dfrac{\pi}{4}$
- B$\dfrac{\pi}{3}$
- Correct answerC$\dfrac{\pi}{2}$
- D$\dfrac{\pi}{6}$
Solution
The intensity $I$ at any point in a Young's double slit experiment is given by $I = 4I_0 \cos^2\left(\dfrac{\phi}{2}\right)$, where $I_0$ is the intensity of each individual wave and $\phi$ is the phase difference between the two interfering waves.
Given the phase difference at point X is $\phi_X = \dfrac{\pi}{3}$, the intensity at X is $I_X = 4I_0 \cos^2\left(\dfrac{\pi/3}{2}\right) = 4I_0 \cos^2\left(\dfrac{\pi}{6}\right) = 4I_0 \left(\dfrac{\sqrt{3}}{2}\right)^2 = 4I_0 \times \dfrac{3}{4} = 3I_0$.
Let the phase difference at point Y be $\phi_Y$. The intensity at Y is $I_Y = 4I_0 \cos^2\left(\dfrac{\phi_Y}{2}\right)$.
The ratio of intensities is given as $\dfrac{I_X}{I_Y} = \dfrac{3}{2}$. Substituting the expressions for $I_X$ and $I_Y$:
$\dfrac{3I_0}{4I_0 \cos^2(\phi_Y/2)} = \dfrac{3}{2}$
$\dfrac{3}{4 \cos^2(\phi_Y/2)} = \dfrac{3}{2}$
$\dfrac{1}{4 \cos^2(\phi_Y/2)} = \dfrac{1}{2}$
$\cos^2\left(\dfrac{\phi_Y}{2}\right) = \dfrac{2}{4} = \dfrac{1}{2}$
$\cos\left(\dfrac{\phi_Y}{2}\right) = \dfrac{1}{\sqrt{2}}$
$\dfrac{\phi_Y}{2} = \dfrac{\pi}{4}$
$\phi_Y = \dfrac{\pi}{2}$.
Answer: $\dfrac{\pi}{2}$
Given the phase difference at point X is $\phi_X = \dfrac{\pi}{3}$, the intensity at X is $I_X = 4I_0 \cos^2\left(\dfrac{\pi/3}{2}\right) = 4I_0 \cos^2\left(\dfrac{\pi}{6}\right) = 4I_0 \left(\dfrac{\sqrt{3}}{2}\right)^2 = 4I_0 \times \dfrac{3}{4} = 3I_0$.
Let the phase difference at point Y be $\phi_Y$. The intensity at Y is $I_Y = 4I_0 \cos^2\left(\dfrac{\phi_Y}{2}\right)$.
The ratio of intensities is given as $\dfrac{I_X}{I_Y} = \dfrac{3}{2}$. Substituting the expressions for $I_X$ and $I_Y$:
$\dfrac{3I_0}{4I_0 \cos^2(\phi_Y/2)} = \dfrac{3}{2}$
$\dfrac{3}{4 \cos^2(\phi_Y/2)} = \dfrac{3}{2}$
$\dfrac{1}{4 \cos^2(\phi_Y/2)} = \dfrac{1}{2}$
$\cos^2\left(\dfrac{\phi_Y}{2}\right) = \dfrac{2}{4} = \dfrac{1}{2}$
$\cos\left(\dfrac{\phi_Y}{2}\right) = \dfrac{1}{\sqrt{2}}$
$\dfrac{\phi_Y}{2} = \dfrac{\pi}{4}$
$\phi_Y = \dfrac{\pi}{2}$.
Answer: $\dfrac{\pi}{2}$
Hence, the correct answer is option: 'C'.