CBSE 2025, Set 2

Question

(a) Find the intensity at a point on the screen in Young's double slit experiment, at which the interfering waves of intensity $I_0$ each, have a path difference of (i) $\frac{\lambda}{3}$ and (ii) $\frac{\lambda}{2}$.

Edvaya · Accepted Answer

The correct answer is Subjective. The phase difference $\Delta\phi$ is related to the path difference $\Delta x$ by the formula:
$$\Delta\phi = \frac{2\pi}{\lambda} \cdot \Delta x$$

The resultant intensity $I$ of two interfering waves, each of intensity $I_0$, is given by:
$$I = 4I_0 \cos^2 \left( \frac{\Delta\phi}{2} ight)$$

**(i) For path difference $\Delta x = \frac{\lambda}{3}$:**
$$\Delta\phi = \frac{2\pi}{\lambda} \cdot \left( \frac{\lambda}{3} ight) = \frac{2\pi}{3} 	ext{ rad}$$
$$I = 4I_0 \cos^2 \left( \frac{2\pi/3}{2} ight) = 4I_0 \cos^2 \left( \frac{\pi}{3} ight)$$
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$:
$$I = 4I_0 \left( \frac{1}{2} ight)^2 = 4I_0 \left( \frac{1}{4} ight) = \boxed{I_0}$$

**(ii) For path difference $\Delta x = \frac{\lambda}{2}$:**
$$\Delta\phi = \frac{2\pi}{\lambda} \cdot \left( \frac{\lambda}{2} ight) = \pi 	ext{ rad}$$
$$I = 4I_0 \cos^2 \left( \frac{\pi}{2} ight)$$
Since $\cos(\frac{\pi}{2}) = 0$:
$$I = 4I_0 (0)^2 = \boxed{0}$$

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