Question

There are three identical point sources $S_1$, $S_2$ and $S_3$ of wavelength $\lambda$ each, are placed near a large screen at $x=d$. If $n_1$, $n_2$ and $n_3$ are number of maxima observed for a pair of sources with third switched off. That means $n_1$ is the number of maxima when $s_1$ switched off; and $n_2$ is the number of maxima when $s_2$ switched off similarly $n_3$ is the number of maxima when $S_3$ switched off. Then $(n_1 + n_2 + n_3)$, for $d = 3\lambda$ and $a = \lambda$, is

• A. 7
• B. 5
• C. 6
• D. 8

Answer 1

Answer: (A)

7

Answer 2

The problem involves calculating the number of maxima observed from pairs of point sources due to interference. The number of maxima for a pair of sources is determined by the range of possible path differences and the condition for constructive interference ($\Delta r = m\lambda$).

1. Pair of sources $S_1$ and $S_2$ ($n_1$) Sources are $S_1=(0, a)$ and $S_2=(0, 0)$. The screen is at $x=d$. The path difference $\Delta r_{12}$ can be approximated for a large screen ($d \gg a$) as $\Delta r_{12} \approx \frac{a^2 - 2ay}{2d}$. For maxima, $\Delta r_{12} = m\lambda$. So, $\frac{a^2 - 2ay}{2d} = m\lambda$. The range of path difference is determined by the limits as $y \to \pm \infty$. As $y \to \pm \infty$, $\Delta r_{12}$ ranges from $-a$ to $a$. So, $-a \le m\lambda \le a$, which implies $-\frac{a}{\lambda} \le m \le \frac{a}{\lambda}$. Given $a = \lambda$, we have $-1 \le m \le 1$. The possible integer values for $m$ are $-1, 0, 1$. Therefore, $n_1 = 3$.

2. Pair of sources $S_1$ and $S_3$ ($n_2$) Sources are $S_1=(0, a)$ and $S_3=(a, 0)$. The screen is at $x=d$. The distance between $S_1$ and $S_3$ is $s = \sqrt{(a-0)^2 + (0-a)^2} = a\sqrt{2}$. The maximum possible path difference between these two sources is equal to the distance between them, which is $a\sqrt{2}$. So, the range of path difference is from $-a\sqrt{2}$ to $a\sqrt{2}$. For maxima, $\Delta r_{13} = m\lambda$. Thus, $-a\sqrt{2} \le m\lambda \le a\sqrt{2}$, which implies $-\frac{a\sqrt{2}}{\lambda} \le m \le \frac{a\sqrt{2}}{\lambda}$. Given $a = \lambda$, we have $-\sqrt{2} \le m \le \sqrt{2}$. The possible integer values for $m$ are $-1, 0, 1$. Therefore, $n_2 = 3$.

3. Pair of sources $S_2$ and $S_3$ ($n_3$) Sources are $S_2=(0, 0)$ and $S_3=(a, 0)$. The screen is at $x=d$. The path difference $\Delta r_{23}$ can be approximated for a large screen as $\Delta r_{23} \approx a - \frac{ay^2}{2d(d-a)}$. The maximum path difference occurs at $y=0$, where $\Delta r_{23}(0) = d - (d-a) = a$. As $|y| \to \infty$, $\Delta r_{23} \to 0$. So, the range of path difference is $(0, a]$. For maxima, $\Delta r_{23} = m\lambda$. Thus, $0 < m\lambda \le a$, which implies $0 < m \le \frac{a}{\lambda}$. Given $a = \lambda$, we have $0 < m \le 1$. The only possible integer value for $m$ is $1$. Therefore, $n_3 = 1$.

The total number of maxima is $(n_1 + n_2 + n_3) = 3 + 3 + 1 = 7$.