CBSE 2025, Set 1

Question

In Young's double slit experimental set-up, the intensity of the central maximum is $I_0$. Calculate the intensity at a point where the path difference between two interfering waves is $\lambda/3$.

Edvaya · Accepted Answer

The correct answer is Subjective. The intensity at any point in the interference pattern is: $$I = I_0 \cos^2\!\left(\frac{\delta}{2}ight)$$ where $\delta$ is the **phase difference** between the two waves at that point. **Converting path difference to phase difference:** $$\delta = \frac{2\pi}{\lambda} 	imes \Delta x = \frac{2\pi}{\lambda} 	imes \frac{\lambda}{3} = \frac{2\pi}{3}$$ **Calculating intensity:** $$I = I_0 \cos^2\!\left(\frac{\delta}{2}ight) = I_0 \cos^2\!\left(\frac{\pi}{3}ight) = I_0 	imes \left(\frac{1}{2}ight)^2 = \boxed{\frac{I_0}{4}}$$ > **Formula reminder:** The general intensity formula is $I = 4I_0'\cos^2(\delta/2)$ where $I_0'$ is the intensity of each individual slit. At central maximum ($\delta = 0$), $I = 4I_0'$, which is given as $I_0$. So $I_0' = I_0/4$. This confirms $I = 4 \cdot \frac{I_0}{4} \cos^2(\pi/3) = I_0 	imes \frac{1}{4} = \frac{I_0}{4}$.

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