Question

Paragraph for Question No. 40 to 42 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 $S_1S_2 = S_2S_3$ = 0.7mm and screen is 30 cm from slits.

40. Find the intensity on the screen at O if $S_1$ and $S_3$ are covered.

• A. $\frac{I_0}{\sqrt{7}}$
• B. $\frac{I_0}{7}$
• C. $\frac{I_0}{\sqrt{6}}$
• D. $\frac{I_0}{6}$

Answer 1

Answer: (B)

$\frac{I_0}{7}$

Answer 2

Let the amplitude of the wave from each identical slit at the central point O on the screen be $a$. Since O is equidistant from all slits and the incident wavefronts are plane, the waves from all slits arrive at O in phase when there are no obstructions or phase-shifting plates.

The intensity is proportional to the square of the resultant amplitude. Let the intensity from a single slit at O be $I_1 \propto a^2$. We can write $I_1 = ka^2$ for some proportionality constant $k$.

When all three slits $S_1, S_2, S_3$ are open, the waves at O are in phase. The resultant amplitude is $A_3 = a + a + a = 3a$. The intensity at O is given as $I_0$. So, $I_0 = k (3a)^2 = 9ka^2 = 9I_1$. Thus, the intensity from a single slit is $I_1 = I_0/9$.

If $S_1$ and $S_3$ are covered, only slit $S_2$ is open. The amplitude at O is $A_1 = a$. The intensity at O is $I_{S_2} = ka^2 = I_1$. Using the relationship $I_1 = I_0/9$, the intensity is $I_{S_2} = I_0/9$.

However, none of the options match the calculated value $I_0/9$. Based on standard physics principles for interference, $I_0/9$ is the correct result. Given the multiple choice format, there might be an error in the options. If forced to choose the closest option, $I_0/7$ is the closest to $I_0/9$.