Q61 / 75 · JEE Main 2026 (06 April Shift 1) · Physics
In interference experiment the path difference between two interfering waves at a point $A$ on the screen is $\lambda/3$, where $\lambda$ is the wavelength of these waves, and at another point $B$ the path difference is $\lambda/6$. The ratio of intensities at points $A$ and $B$ is _______.
- A$3$
- B$4$
- Correct answerC$1/3$
- D$1/4$
Solution
The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the relation $\Delta \phi = \dfrac{2\pi}{\lambda} \Delta x$.
For point $A$, the path difference is $\Delta x_A = \dfrac{\lambda}{3}$.
$\Delta \phi_A = \dfrac{2\pi}{\lambda} \times \dfrac{\lambda}{3} = \dfrac{2\pi}{3}$
The intensity at point $A$ is given by $I_A = I_{max} \cos^2\left(\dfrac{\Delta \phi_A}{2}\right)$.
$I_A = I_{max} \cos^2\left(\dfrac{\pi}{3}\right) = I_{max} \left(\dfrac{1}{2}\right)^2 = \dfrac{I_{max}}{4}$
For point $B$, the path difference is $\Delta x_B = \dfrac{\lambda}{6}$.
$\Delta \phi_B = \dfrac{2\pi}{\lambda} \times \dfrac{\lambda}{6} = \dfrac{\pi}{3}$
The intensity at point $B$ is given by $I_B = I_{max} \cos^2\left(\dfrac{\Delta \phi_B}{2}\right)$.
$I_B = I_{max} \cos^2\left(\dfrac{\pi}{6}\right) = I_{max} \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3I_{max}}{4}$
The ratio of intensities at points $A$ and $B$ is:
$\dfrac{I_A}{I_B} = \dfrac{\dfrac{I_{max}}{4}}{\dfrac{3I_{max}}{4}} = \dfrac{1}{3}$
Answer: $1/3$
For point $A$, the path difference is $\Delta x_A = \dfrac{\lambda}{3}$.
$\Delta \phi_A = \dfrac{2\pi}{\lambda} \times \dfrac{\lambda}{3} = \dfrac{2\pi}{3}$
The intensity at point $A$ is given by $I_A = I_{max} \cos^2\left(\dfrac{\Delta \phi_A}{2}\right)$.
$I_A = I_{max} \cos^2\left(\dfrac{\pi}{3}\right) = I_{max} \left(\dfrac{1}{2}\right)^2 = \dfrac{I_{max}}{4}$
For point $B$, the path difference is $\Delta x_B = \dfrac{\lambda}{6}$.
$\Delta \phi_B = \dfrac{2\pi}{\lambda} \times \dfrac{\lambda}{6} = \dfrac{\pi}{3}$
The intensity at point $B$ is given by $I_B = I_{max} \cos^2\left(\dfrac{\Delta \phi_B}{2}\right)$.
$I_B = I_{max} \cos^2\left(\dfrac{\pi}{6}\right) = I_{max} \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3I_{max}}{4}$
The ratio of intensities at points $A$ and $B$ is:
$\dfrac{I_A}{I_B} = \dfrac{\dfrac{I_{max}}{4}}{\dfrac{3I_{max}}{4}} = \dfrac{1}{3}$
Answer: $1/3$
Hence, the correct answer is option: 'C'.