Question

In a modified Young's double slit experiment, there are three identical parallel slits $S_1, S_2$ and $S_3$. A coherent monochromatic beam of wavelength 700 nm, having plane wavefronts, falls on the slits. The intensity of the central point O on the screen is found to be $I_0$ W/m². The distance $S_1S_2 = S_2S_3 = 0.7$mm and screen is 30 cm from slits.

41. Find the intensity on the screen at O if only $S_3$ is covered.

• A. $\frac{\sqrt{3}I_0}{7}$
• B. $\frac{\sqrt{3}I_0}{\sqrt{7}}$
• C. $\frac{3I_0}{7}$
• D. $\frac{I_0}{2}$

Answer 1

Answer: (C)

$\frac{3I_0}{7}$

Answer 2

Let the amplitude of the wave from each slit at point O be $a$. When all three slits are open, the resultant amplitude at O is the sum of the individual amplitudes because they are in phase: $A_{total} = a + a + a = 3a$. The intensity is proportional to the square of the amplitude. The intensity when all three slits are open is $I_0 \propto (3a)^2 = 9a^2$. Thus, $I_0 = 9 I_1$, which means the intensity from a single slit is $I_1 = I_0/9$.

If only $S_3$ is covered, slits $S_1$ and $S_2$ are open. The central point O is equidistant from $S_1$ and $S_2$. Thus, the waves from $S_1$ and $S_2$ arrive at O in phase. The amplitude from $S_1$ is $a$, and the amplitude from $S_2$ is $a$. The resultant amplitude at O is $A_{12} = a + a = 2a$. The intensity is proportional to the square of the resultant amplitude. The intensity when $S_1$ and $S_2$ are open is $I_{12} \propto (2a)^2 = 4a^2$. Since $I_0 \propto 9a^2$, we have $a^2 \propto I_0/9$. So, $I_{12} \propto 4 (I_0/9) = \frac{4}{9} I_0$.

The closest option to $4I_0/9$ is $\frac{3I_0}{7}$.