Question

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
• C. 1/3
• D. 1/4

Hint:

Relate the path difference to the phase difference using $\phi = \frac{2\pi}{\lambda}\Delta x$, then use the formula for resultant intensity in interference $I = 4I_0 \cos^2(\phi/2)$.

Answer 1

The Correct Option is C

The intensity $I$ of light resulting from the interference of two waves of equal intensity $I_0$ is given by the formula:
$$ I = 4I_0 \cos^2\left(\frac{\phi}{2}\right) $$
where $\phi$ is the phase difference between the two waves. The phase difference is related to the path difference $\Delta x$ by the equation:
$$ \phi = \frac{2\pi}{\lambda} \Delta x $$
For point $A$, the path difference is $\Delta x_A = \lambda/3$. The phase difference is:
$$ \phi_A = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3} $$
The intensity at point $A$ is:
$$ I_A = 4I_0 \cos^2\left(\frac{2\pi/3}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{3}\right) = 4I_0 \left(\frac{1}{2}\right)^2 = I_0 $$
For point $B$, the path difference is $\Delta x_B = \lambda/6$. The phase difference is:
$$ \phi_B = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{6} = \frac{\pi}{3} $$
The intensity at point $B$ is:
$$ I_B = 4I_0 \cos^2\left(\frac{\pi/3}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{6}\right) = 4I_0 \left(\frac{\sqrt{3}}{2}\right)^2 = 4I_0 \cdot \frac{3}{4} = 3I_0 $$
The ratio of the intensities at points $A$ and $B$ is:
$$ \frac{I_A}{I_B} = \frac{I_0}{3I_0} = \frac{1}{3} $$