Question

Find the intensity of light at a point on the screen when two interfering waves of the same intensity ($I_0$) have a path difference of (i) $\frac{\lambda}{4}$ and (ii) $\frac{\lambda}{3}$.

Hint:

Remember key intensity values for equal source intensities $I_0$:
$\Delta x = 0 \implies I = 4I_0$ (Central Maxima)
$\Delta x = \lambda/4 \implies I = 2I_0$
$\Delta x = \lambda/3 \implies I = I_0$
$\Delta x = \lambda/2 \implies I = 0$ (Minima)

Answer 1

Concept:
• The relation between phase difference $\phi$ and path difference $\Delta x$ is $\phi = \frac{2\pi}{\lambda} \cdot \Delta x$.

• Resultant intensity of two coherent waves each of intensity $I_0$ is $I = 4 I_0 \cos^2\left(\frac{\phi}{2}\right)$.

Step 1: Formulas Used
Phase difference $\phi$ corresponding to path difference $\Delta x$:
\[ \phi = \left( \frac{2\pi}{\lambda} \right) \Delta x \]
Resultant intensity $I$:
\[ I = I_0 + I_0 + 2\sqrt{I_0 I_0}\cos \phi = 2 I_0 (1 + \cos \phi) = 4 I_0 \cos^2\left(\frac{\phi}{2}\right) \]

Step 2: Case (i): Path Difference $\Delta x = \frac{\lambda}{4}$
Calculate phase difference $\phi_1$:
\[ \phi_1 = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2} \text{ radians} \]
Calculate intensity $I_1$:
\[ I_1 = 4 I_0 \cos^2\left( \frac{\pi/2}{2} \right) = 4 I_0 \cos^2\left(\frac{\pi}{4}\right) \]
Since $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$:
\[ I_1 = 4 I_0 \left(\frac{1}{\sqrt{2}}\right)^2 = 4 I_0 \left(\frac{1}{2}\right) = 2 I_0 \]

Step 3: Case (ii): Path Difference $\Delta x = \frac{\lambda}{3}$
Calculate phase difference $\phi_2$:
\[ \phi_2 = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3} \text{ radians} \]
Calculate intensity $I_2$:
\[ I_2 = 4 I_0 \cos^2\left( \frac{2\pi/3}{2} \right) = 4 I_0 \cos^2\left(\frac{\pi}{3}\right) \]
Since $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$:
\[ I_2 = 4 I_0 \left(\frac{1}{2}\right)^2 = 4 I_0 \left(\frac{1}{4}\right) = I_0 \]
let's verify:
$\cos\left(\frac{2\pi}{3}\right) = -1/2$.
$I_2 = 2 I_0 (1 + \cos(2\pi/3)) = 2 I_0 (1 - 0.5) = 2 I_0 \times 0.5 = I_0$.
Or using $4 I_0 \cos^2(\pi/3) = 4 I_0 (1/2)^2 = I_0$.

Step 4: Conclusion
(i) For path difference $\frac{\lambda}{4}$, resultant intensity is $2 I_0$.
(ii) For path difference $\frac{\lambda}{3}$, resultant intensity is $I_0$.