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

In any interference pattern where the source intensities are equal, the resultant intensity is proportional to the square of the cosine of half the phase difference:
$$ I \propto \cos^2\left(\frac{\phi}{2}\right) $$
The phase difference $\phi$ is proportional to the path difference $\Delta x$ as $\phi = (2\pi/\lambda)\Delta x$. Thus, we can write:
$$ \frac{I_A}{I_B} = \frac{\cos^2(\phi_A/2)}{\cos^2(\phi_B/2)} $$
Step 1: Calculate the phase difference at point $A$ and $B$.
$\phi_A = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3} = 120^\circ$
$\phi_B = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{6} = \frac{\pi}{3} = 60^\circ$
Step 2: Substitute these values into the intensity ratio formula.
$$ \frac{I_A}{I_B} = \frac{\cos^2(120^\circ/2)}{\cos^2(60^\circ/2)} = \frac{\cos^2(60^\circ)}{\cos^2(30^\circ)} $$
Step 3: Evaluate the trigonometric functions.
$\cos(60^\circ) = 1/2 \implies \cos^2(60^\circ) = 1/4$
$\cos(30^\circ) = \sqrt{3}/2 \implies \cos^2(30^\circ) = 3/4$
$$ \frac{I_A}{I_B} = \frac{1/4}{3/4} = \frac{1}{3} $$
The ratio of intensities at points $A$ and $B$ is $1/3$.