CBSE 2025, Set 3

Question

In a Young's double-slit experiment, the intensity at the central maximum in the interference pattern on the screen is $I_0$. Find the intensity at a point on the screen where the path difference between the interfering waves is $\frac{\lambda}{6}$.

Edvaya · Accepted Answer

The correct answer is Subjective. The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula:

$$\phi = \frac{2\pi}{\lambda} 	imes \Delta x$$

Given the path difference $\Delta x = \frac{\lambda}{6}$, we substitute this into the equation:

$$\phi = \frac{2\pi}{\lambda} 	imes \frac{\lambda}{6} = \frac{\pi}{3}	ext{ rad}$$

The resultant intensity $I$ in Young's double-slit experiment is given by:

$$I = I_0 \cos^2 \left( \frac{\phi}{2} ight)$$

Substitute the value of $\phi$:

$$I = I_0 \cos^2 \left( \frac{\pi / 3}{2} ight) = I_0 \cos^2 \left( \frac{\pi}{6} ight)$$

Since $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$:

$$I = I_0 \left( \frac{\sqrt{3}}{2} ight)^2$$

$$I = \boxed{\frac{3I_0}{4}}$$

> For two identical slits, max intensity is $4I_{slit}$, which is defined here as $I_0$.

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