OPTICS OBJECTIVE 10

1. Path Difference Formulation

This setup is analogous to Lloyd’s Mirror. Interference occurs between the direct wave from the satellite and the wave reflected from the lake surface. For a receiver at height $h$ and satellite elevation angle $\theta$, the path difference $\Delta x$ is approximately:

$$ \Delta x \approx 2h \sin \theta $$

2. Condition for Maxima

We are given that maxima occur at $\theta_1 = 3^\circ$ and the next maxima at $\theta_2 = 6^\circ$.

Let the path difference condition for the $m$-th maxima be $\Delta x_m$. The difference in path difference between two consecutive maxima corresponds to one full wavelength $\lambda$.

$$ \Delta x_2 – \Delta x_1 = \lambda $$ $$ 2h \sin \theta_2 – 2h \sin \theta_1 = \lambda $$ $$ \lambda = 2h (\sin 6^\circ – \sin 3^\circ) $$

3. Calculation

Given $h = 4.0 \, \text{m} = 400 \, \text{cm}$.

Using values: $\sin 6^\circ \approx 0.1045$ and $\sin 3^\circ \approx 0.0523$.

$$ \lambda = 2(400) (0.1045 – 0.0523) $$ $$ \lambda = 800 (0.0522) $$ $$ \lambda = 41.76 \, \text{cm} $$

Rounding to the nearest integer, we get 42 cm.