Find Integer Parameter in Young's Double Slit Water Immersion Maxima

To find the value of $p$, we analyze the optical path difference between the light rays arriving at a point $P$ on the water surface from the two slits $S_1$ and $S_2$.

1. Geometric Setup: Let the plane containing the two slits $S_1$ and $S_2$ lie along the vertical $y$-axis, with the origin $(0, 0)$ located at the midpoint between the slits on the water surface:

• Slit $S_1$ is located at $(0, d)$ in air (refractive index $\mu_1 = 1$).
• Slit $S_2$ is located at $(0, -d)$ in water (refractive index $\mu_2 = \frac{4}{3}$).
• A point $P$ on the water surface is at a distance $x$ from the origin, so its coordinates are $(x, 0)$.

2. Geometric Path Lengths: The geometric distance from $S_1$ to $P$ is: $S_1P = \sqrt{x^2 + d^2}$

The geometric distance from $S_2$ to $P$ is: $S_2P = \sqrt{x^2 + (-d)^2} = \sqrt{x^2 + d^2}$

3. Optical Path Lengths:

• The ray from $S_1$ travels entirely through air to reach point $P$: $\text{Optical Path } (L_1) = \mu_1 \cdot S_1P = 1 \cdot \sqrt{x^2 + d^2}$

• The ray from $S_2$ travels entirely through water to reach point $P$: $\text{Optical Path } (L_2) = \mu_2 \cdot S_2P = \frac{4}{3} \cdot \sqrt{x^2 + d^2}$

4. Optical Path Difference and Maxima Condition: The optical path difference $\Delta L$ between the two rays at point $P$ is given by: $\Delta L = L_2 - L_1 = \left(\frac{4}{3} - 1\right) \sqrt{x^2 + d^2} = \frac{1}{3} \sqrt{x^2 + d^2}$

For constructive interference (maxima), the optical path difference must be an integer multiple of the wavelength $\lambda$ of light in air: $\Delta L = m \lambda \quad (m \in \mathbb{Z})$

Substituting $\Delta L$: $\frac{1}{3} \sqrt{x^2 + d^2} = m \lambda$

$\sqrt{x^2 + d^2} = 3 m \lambda$

Squaring both sides gives: $x^2 + d^2 = 9 m^2 \lambda^2$

$x^2 = 9 m^2 \lambda^2 - d^2$

5. Comparing with the Given Relation: The problem gives the positions of maxima as: $x^2 = p^2 m^2 \lambda^2 - d^2$

Comparing the coefficients of $m^2 \lambda^2$: $p^2 = 9 \implies p = 3$