Question

On the ‘+y’ axis, two identical coherent sources S1 and S2 , separated by distance ‘d’ emits light wave each of wave length ' ' 4000   nm. Plane x=0 serves as an imaginary boundary between two medium of refractive index  1 and  1.125. If a maxima is observed just on the left of origin and a minima is observe just on the right of origin, then the minimum non-zero value of d (in m) is 2x . Then x is

• A. 8
• B. 4
• C. 16
• D. 2

Answer 1

Answer: (A)

8

Answer 2

Let the geometric path difference between the two sources be $\Delta r$.

For constructive interference (maxima) in the first medium ($\mu_1 = 1$): $\Delta L_1 = \mu_1 \Delta r = m\lambda$ $1 \cdot \Delta r = m\lambda$ $\Delta r = m\lambda$, where $m$ is an integer.

For destructive interference (minima) in the second medium ($\mu_2 = 1.125 = \frac{9}{8}$): $\Delta L_2 = \mu_2 \Delta r = (m' + \frac{1}{2})\lambda$ $\frac{9}{8} \Delta r = (m' + \frac{1}{2})\lambda$, where $m'$ is an integer.

Substituting $\Delta r = m\lambda$ from the first condition into the second: $\frac{9}{8} (m\lambda) = (m' + \frac{1}{2})\lambda$ $\frac{9m}{8} = m' + \frac{1}{2}$ $9m = 8m' + 4$

We need to find the minimum non-zero integer values for $m$ and $m'$ that satisfy this equation. Let's test values for $m$: If $m=1$, $9 = 8m' + 4 \implies 8m' = 5$ (no integer solution for $m'$). If $m=2$, $18 = 8m' + 4 \implies 8m' = 14$ (no integer solution for $m'$). If $m=3$, $27 = 8m' + 4 \implies 8m' = 23$ (no integer solution for $m'$). If $m=4$, $36 = 8m' + 4 \implies 8m' = 32 \implies m' = 4$.

The minimum non-zero integer value for $m$ is 4. This corresponds to a minimum non-zero geometric path difference: $\Delta r = m\lambda = 4\lambda$

The problem states that the sources are separated by a distance $d$, and we are looking for the minimum non-zero value of $d$. In such interference problems, the separation $d$ is often directly related to the geometric path difference. The simplest assumption that satisfies the conditions is that the geometric path difference $\Delta r$ is equal to the source separation $d$. So, $d = \Delta r = 4\lambda$.

Given $\lambda = 4000$ nm $= 4 \mu$m. $d = 4 \times (4 \mu\text{m}) = 16 \mu$m.

The problem also states that the minimum non-zero value of $d$ is $2x$. $2x = 16 \mu\text{m}$ $x = \frac{16}{2}$ $x = 8$